6月26日
被1226煩得頭都大了，從下午斷斷續續調到晚。最后數組開小給報個Tle看了半天沒看出來...
6月30日
每日補題，進度停滯中...
7月10日
第一階段結束...不做鴿子了不做鴿子了。
7月11日
被卡了....還沒題解....生命--。
7月21日
達到40題之后已經咸魚了。隨緣做做。
hdu1219
遍歷計數。
1 #include<bits/stdc++.h> 2 #define QAQ 0 3 4 using namespace std; 5 6 char a[100010]; 7 int cnt[200]; 8 int main(){ 9 while(cin.getline(a,100005)){ 10 memset(cnt,0,sizeof(cnt)); 11 int l=strlen(a); 12 for(int i=0;i<l;i++){ 13 cnt[a[i]]++; 14 } 15 for(int i='a';i<='z';i++){ 16 printf("%c:%d\n",i,cnt[i]); 17 } 18 printf("\n"); 19 } 20 return QAQ; 21 }
hdu1220
總對數減去相鄰的對數。相鄰的對數分類為 頂點、棱、面、體內 的小正方體，分別對應鄰接3、4、5、6個，每對算了兩次。
1 #include<bits/stdc++.h> 2 #define QAQ 0 3 4 using namespace std; 5 typedef long long ll; 6 7 ll n; 8 9 int main(){ 10 while(scanf("%lld",&n)!=EOF){ 11 if(n==1){ 12 printf("%lld\n",0); 13 continue; 14 } 15 ll sum=n*n*n*(n*n*n-1)/2,cnt; 16 cnt=8*3 + 12*(n-2)*4 + 6*(n-2)*(n-2)*5 + (n-2)*(n-2)*(n-2)*6; 17 printf("%lld\n",sum-cnt/2); 18 } 19 return QAQ; 20 }
hdu1221
最大值一定到在圓心到矩形四個頂點里，最小值除了到頂點距離外，再判一下圓心到四條邊距離。
1 #include<bits/stdc++.h> 2 #define QAQ 0 3 4 using namespace std; 5 typedef long long ll; 6 const double eps=1e-7; 7 8 double x,y,r,a1,b1,a2,b2,a3,b3,a4,b4,maxi,mini; 9 ll T; 10 11 double dis(double x1, double y1, double x2, double y2){ 12 return sqrt( (x1-x2)*(x1-x2) + (y1-y2)*(y1-y2) ); 13 } 14 15 int main(){ 16 scanf("%lld",&T); 17 while(T--){ 18 scanf("%lf%lf%lf%lf%lf%lf%lf",&x,&y,&r,&a1,&b1,&a2,&b2); 19 a3=a1; b3=b2; a4=a2; b4=b1; 20 double l1=dis(x,y,a1,b1),l2=dis(x,y,a2,b2), 21 l3=dis(x,y,a3,b3),l4=dis(x,y,a4,b4); 22 23 maxi=max( max(l1,l2) , max(l3,l4) ); 24 mini=min( min(l1,l2) , min(l3,l4) ); 25 26 if(x>a1 && x<a2){ 27 mini=min( mini, min( fabs(y-b1) , fabs(y-b2) ) ); 28 } 29 30 if(y>b1 && y<b2){ 31 mini=min( mini, min( fabs(x-a1) , fabs(x-a2) ) ); 32 } 33 34 if(mini<=r && maxi>=r){ 35 printf("YES"); 36 } 37 else printf("NO"); 38 printf("\n"); 39 } 40 return QAQ; 41 }
hdu1222
如果m和n的最大公約數不為1，則一定有安全位置。
1 #include<bits/stdc++.h> 2 #define QAQ 0 3 4 using namespace std; 5 6 int T,m,n; 7 8 int gcd(int a,int b){ 9 return b==0?a:gcd(b,a%b); 10 } 11 12 int main(){ 13 scanf("%d",&T); 14 while(T--){ 15 scanf("%d%d",&m,&n); 16 if(gcd(m,n)!=1) printf("YES"); 17 else printf("NO"); 18 printf("\n"); 19 } 20 return QAQ; 21 }
hdu1223
有等號連接的為一個塊，dp[i][j]表示i個字母有j個塊。對於i-1個字母分兩類：1.原有j塊，即添加等號，顯然和每個塊相等為一種，共j種；2.原有j-1塊，即添加小於號，j-1個塊有j個空位可以放入，也為j種。所以dp[i][j]=dp[i-1][j]*j+dp[i-1]][j-1]*j。
本題沒有mod，要用大數。感謝顏神的模板支援o(*￣▽￣*)ブ。
1 #include<bits/stdc++.h> 2 #define QAQ 0 3 4 using namespace std; 5 typedef long long ll; 6 7 int T,n; 8 9 const int base = 1000000000; 10 const int base_digits = 9; 11 12 struct Bigint { 13 vector<int> z; 14 int sign; 15 16 Bigint() : 17 sign(1) { 18 } 19 20 Bigint(long long v) { 21 *this = v; 22 } 23 24 Bigint(const string &s) { 25 read(s); 26 } 27 28 void operator=(const Bigint &v) { 29 sign = v.sign; 30 z = v.z; 31 } 32 33 void operator=(long long v) { 34 sign = 1; 35 if (v < 0) 36 sign = -1, v = -v; 37 z.clear(); 38 for (; v > 0; v = v / base) 39 z.push_back(v % base); 40 } 41 42 Bigint operator+(const Bigint &v) const { 43 if (sign == v.sign) { 44 Bigint res = v; 45 46 for (int i = 0, carry = 0; i < (int) max(z.size(), v.z.size()) || carry; ++i) { 47 if (i == (int) res.z.size()) 48 res.z.push_back(0); 49 res.z[i] += carry + (i < (int) z.size() ? z[i] : 0); 50 carry = res.z[i] >= base; 51 if (carry) 52 res.z[i] -= base; 53 } 54 return res; 55 } 56 return *this - (-v); 57 } 58 59 Bigint operator-(const Bigint &v) const { 60 if (sign == v.sign) { 61 if (abs() >= v.abs()) { 62 Bigint res = *this; 63 for (int i = 0, carry = 0; i < (int) v.z.size() || carry; ++i) { 64 res.z[i] -= carry + (i < (int) v.z.size() ? v.z[i] : 0); 65 carry = res.z[i] < 0; 66 if (carry) 67 res.z[i] += base; 68 } 69 res.trim(); 70 return res; 71 } 72 return -(v - *this); 73 } 74 return *this + (-v); 75 } 76 77 void operator*=(int v) { 78 if (v < 0) 79 sign = -sign, v = -v; 80 for (int i = 0, carry = 0; i < (int) z.size() || carry; ++i) { 81 if (i == (int) z.size()) 82 z.push_back(0); 83 long long cur = z[i] * (long long) v + carry; 84 carry = (int) (cur / base); 85 z[i] = (int) (cur % base); 86 //asm("divl %%ecx" : "=a"(carry), "=d"(a[i]) : "A"(cur), "c"(base)); 87 } 88 trim(); 89 } 90 91 Bigint operator*(int v) const { 92 Bigint res = *this; 93 res *= v; 94 return res; 95 } 96 97 friend pair<Bigint, Bigint> divmod(const Bigint &a1, const Bigint &b1) { 98 int norm = base / (b1.z.back() + 1); 99 Bigint a = a1.abs() * norm; 100 Bigint b = b1.abs() * norm; 101 Bigint q, r; 102 q.z.resize(a.z.size()); 103 104 for (int i = a.z.size() - 1; i >= 0; i--) { 105 r *= base; 106 r += a.z[i]; 107 int s1 = b.z.size() < r.z.size() ? r.z[b.z.size()] : 0; 108 int s2 = b.z.size() - 1 < r.z.size() ? r.z[b.z.size() - 1] : 0; 109 int d = ((long long) s1 * base + s2) / b.z.back(); 110 r -= b * d; 111 while (r < 0) 112 r += b, --d; 113 q.z[i] = d; 114 } 115 116 q.sign = a1.sign * b1.sign; 117 r.sign = a1.sign; 118 q.trim(); 119 r.trim(); 120 return make_pair(q, r / norm); 121 } 122 123 friend Bigint sqrt(const Bigint &a1) { 124 Bigint a = a1; 125 while (a.z.empty() || a.z.size() % 2 == 1) 126 a.z.push_back(0); 127 128 int n = a.z.size(); 129 130 int firstDigit = (int) sqrt((double) a.z[n - 1] * base + a.z[n - 2]); 131 int norm = base / (firstDigit + 1); 132 a *= norm; 133 a *= norm; 134 while (a.z.empty() || a.z.size() % 2 == 1) 135 a.z.push_back(0); 136 137 Bigint r = (long long) a.z[n - 1] * base + a.z[n - 2]; 138 firstDigit = (int) sqrt((double) a.z[n - 1] * base + a.z[n - 2]); 139 int q = firstDigit; 140 Bigint res; 141 142 for(int j = n / 2 - 1; j >= 0; j--) { 143 for(; ; --q) { 144 Bigint r1 = (r - (res * 2 * base + q) * q) * base * base + (j > 0 ? (long long) a.z[2 * j - 1] * base + a.z[2 * j - 2] : 0); 145 if (r1 >= 0) { 146 r = r1; 147 break; 148 } 149 } 150 res *= base; 151 res += q; 152 153 if (j > 0) { 154 int d1 = res.z.size() + 2 < r.z.size() ? r.z[res.z.size() + 2] : 0; 155 int d2 = res.z.size() + 1 < r.z.size() ? r.z[res.z.size() + 1] : 0; 156 int d3 = res.z.size() < r.z.size() ? r.z[res.z.size()] : 0; 157 q = ((long long) d1 * base * base + (long long) d2 * base + d3) / (firstDigit * 2); 158 } 159 } 160 161 res.trim(); 162 return res / norm; 163 } 164 165 Bigint operator/(const Bigint &v) const { 166 return divmod(*this, v).first; 167 } 168 169 Bigint operator%(const Bigint &v) const { 170 return divmod(*this, v).second; 171 } 172 173 void operator/=(int v) { 174 if (v < 0) 175 sign = -sign, v = -v; 176 for (int i = (int) z.size() - 1, rem = 0; i >= 0; --i) { 177 long long cur = z[i] + rem * (long long) base; 178 z[i] = (int) (cur / v); 179 rem = (int) (cur % v); 180 } 181 trim(); 182 } 183 184 Bigint operator/(int v) const { 185 Bigint res = *this; 186 res /= v; 187 return res; 188 } 189 190 int operator%(int v) const { 191 if (v < 0) 192 v = -v; 193 int m = 0; 194 for (int i = z.size() - 1; i >= 0; --i) 195 m = (z[i] + m * (long long) base) % v; 196 return m * sign; 197 } 198 199 void operator+=(const Bigint &v) { 200 *this = *this + v; 201 } 202 void operator-=(const Bigint &v) { 203 *this = *this - v; 204 } 205 void operator*=(const Bigint &v) { 206 *this = *this * v; 207 } 208 void operator/=(const Bigint &v) { 209 *this = *this / v; 210 } 211 212 bool operator<(const Bigint &v) const { 213 if (sign != v.sign) 214 return sign < v.sign; 215 if (z.size() != v.z.size()) 216 return z.size() * sign < v.z.size() * v.sign; 217 for (int i = z.size() - 1; i >= 0; i--) 218 if (z[i] != v.z[i]) 219 return z[i] * sign < v.z[i] * sign; 220 return false; 221 } 222 223 bool operator>(const Bigint &v) const { 224 return v < *this; 225 } 226 bool operator<=(const Bigint &v) const { 227 return !(v < *this); 228 } 229 bool operator>=(const Bigint &v) const { 230 return !(*this < v); 231 } 232 bool operator==(const Bigint &v) const { 233 return !(*this < v) && !(v < *this); 234 } 235 bool operator!=(const Bigint &v) const { 236 return *this < v || v < *this; 237 } 238 239 void trim() { 240 while (!z.empty() && z.back() == 0) 241 z.pop_back(); 242 if (z.empty()) 243 sign = 1; 244 } 245 246 bool isZero() const { 247 return z.empty() || (z.size() == 1 && !z[0]); 248 } 249 250 Bigint operator-() const { 251 Bigint res = *this; 252 res.sign = -sign; 253 return res; 254 } 255 256 Bigint abs() const { 257 Bigint res = *this; 258 res.sign *= res.sign; 259 return res; 260 } 261 262 long long longValue() const { 263 long long res = 0; 264 for (int i = z.size() - 1; i >= 0; i--) 265 res = res * base + z[i]; 266 return res * sign; 267 } 268 269 friend Bigint gcd(const Bigint &a, const Bigint &b) { 270 return b.isZero() ? a : gcd(b, a % b); 271 } 272 friend Bigint lcm(const Bigint &a, const Bigint &b) { 273 return a / gcd(a, b) * b; 274 } 275 276 void read(const string &s) { 277 sign = 1; 278 z.clear(); 279 int pos = 0; 280 while (pos < (int) s.size() && (s[pos] == '-' || s[pos] == '+')) { 281 if (s[pos] == '-') 282 sign = -sign; 283 ++pos; 284 } 285 for (int i = s.size() - 1; i >= pos; i -= base_digits) { 286 int x = 0; 287 for (int j = max(pos, i - base_digits + 1); j <= i; j++) 288 x = x * 10 + s[j] - '0'; 289 z.push_back(x); 290 } 291 trim(); 292 } 293 294 friend istream& operator>>(istream &stream, Bigint &v) { 295 string s; 296 stream >> s; 297 v.read(s); 298 return stream; 299 } 300 301 friend ostream& operator<<(ostream &stream, const Bigint &v) { 302 if (v.sign == -1) 303 stream << '-'; 304 stream << (v.z.empty() ? 0 : v.z.back()); 305 for (int i = (int) v.z.size() - 2; i >= 0; --i) 306 stream << setw(base_digits) << setfill('0') << v.z[i]; 307 return stream; 308 } 309 310 static vector<int> convert_base(const vector<int> &a, int old_digits, int new_digits) { 311 vector<long long> p(max(old_digits, new_digits) + 1); 312 p[0] = 1; 313 for (int i = 1; i < (int) p.size(); i++) 314 p[i] = p[i - 1] * 10; 315 vector<int> res; 316 long long cur = 0; 317 int cur_digits = 0; 318 for (int i = 0; i < (int) a.size(); i++) { 319 cur += a[i] * p[cur_digits]; 320 cur_digits += old_digits; 321 while (cur_digits >= new_digits) { 322 res.push_back(int(cur % p[new_digits])); 323 cur /= p[new_digits]; 324 cur_digits -= new_digits; 325 } 326 } 327 res.push_back((int) cur); 328 while (!res.empty() && res.back() == 0) 329 res.pop_back(); 330 return res; 331 } 332 333 typedef vector<long long> vll; 334 335 static vll karatsubaMultiply(const vll &a, const vll &b) { 336 int n = a.size(); 337 vll res(n + n); 338 if (n <= 32) { 339 for (int i = 0; i < n; i++) 340 for (int j = 0; j < n; j++) 341 res[i + j] += a[i] * b[j]; 342 return res; 343 } 344 345 int k = n >> 1; 346 vll a1(a.begin(), a.begin() + k); 347 vll a2(a.begin() + k, a.end()); 348 vll b1(b.begin(), b.begin() + k); 349 vll b2(b.begin() + k, b.end()); 350 351 vll a1b1 = karatsubaMultiply(a1, b1); 352 vll a2b2 = karatsubaMultiply(a2, b2); 353 354 for (int i = 0; i < k; i++) 355 a2[i] += a1[i]; 356 for (int i = 0; i < k; i++) 357 b2[i] += b1[i]; 358 359 vll r = karatsubaMultiply(a2, b2); 360 for (int i = 0; i < (int) a1b1.size(); i++) 361 r[i] -= a1b1[i]; 362 for (int i = 0; i < (int) a2b2.size(); i++) 363 r[i] -= a2b2[i]; 364 365 for (int i = 0; i < (int) r.size(); i++) 366 res[i + k] += r[i]; 367 for (int i = 0; i < (int) a1b1.size(); i++) 368 res[i] += a1b1[i]; 369 for (int i = 0; i < (int) a2b2.size(); i++) 370 res[i + n] += a2b2[i]; 371 return res; 372 } 373 374 Bigint operator*(const Bigint &v) const { 375 vector<int> a6 = convert_base(this->z, base_digits, 6); 376 vector<int> b6 = convert_base(v.z, base_digits, 6); 377 vll a(a6.begin(), a6.end()); 378 vll b(b6.begin(), b6.end()); 379 while (a.size() < b.size()) 380 a.push_back(0); 381 while (b.size() < a.size()) 382 b.push_back(0); 383 while (a.size() & (a.size() - 1)) 384 a.push_back(0), b.push_back(0); 385 vll c = karatsubaMultiply(a, b); 386 Bigint res; 387 res.sign = sign * v.sign; 388 for (int i = 0, carry = 0; i < (int) c.size(); i++) { 389 long long cur = c[i] + carry; 390 res.z.push_back((int) (cur % 1000000)); 391 carry = (int) (cur / 1000000); 392 } 393 res.z = convert_base(res.z, 6, base_digits); 394 res.trim(); 395 return res; 396 } 397 }dp[60][60],sum[60]; 398 399 int main(){ 400 scanf("%d",&T); 401 for(int i=1;i<=50;i++){ 402 dp[i][1]=1; 403 } 404 for(int i=2;i<=50;i++){ 405 for(int j=2;j<=i;j++){ 406 dp[i][j]=dp[i-1][j]*j + dp[i-1][j-1]*j; 407 } 408 } 409 for(int i=1;i<=50;i++){ 410 for(int j=1;j<=i;j++){ 411 sum[i]=sum[i]+dp[i][j]; 412 } 413 } 414 while(T--){ 415 scanf("%d",&n); 416 cout<<sum[n]<<'\n'; 417 } 418 return QAQ; 419 }
hdu1224
數據很小，dfs一遍，順帶記錄一下走的路線。
1 #include<bits/stdc++.h> 2 #define QAQ 0 3 4 using namespace std; 5 6 int T,n,cas,maxi,m,u,v,a[102]; 7 vector<int>ve[102],tmp,ans; 8 9 void dfs(int x,int t){ 10 if(x==n+1){ 11 if(t>maxi){ 12 maxi=t; 13 ans.clear(); 14 ans=tmp; 15 } 16 return; 17 } 18 for(auto y:ve[x]){ 19 tmp.push_back(x); 20 dfs(y,t+a[y]); 21 tmp.pop_back(); 22 } 23 } 24 25 int main(){ 26 scanf("%d",&T); 27 while(T--){ 28 maxi=-1; 29 tmp.clear(); ans.clear(); 30 scanf("%d",&n); 31 for(int i=1;i<=n;i++){ 32 scanf("%d",&a[i]); 33 ve[i].clear(); 34 } 35 a[n+1]=a[1]; 36 scanf("%d",&m); 37 for(int i=1;i<=m;i++){ 38 scanf("%d%d",&u,&v); 39 ve[u].push_back(v); 40 } 41 dfs(1,a[1]); 42 cas++; 43 printf("CASE %d#\npoints : %d\ncircuit : ",cas,maxi); 44 for(auto x:ans){ 45 printf("%d->",x); 46 } 47 printf("1\n"); 48 if(T) printf("\n"); 49 } 50 return QAQ; 51 }
hdu1225
按題意模擬，用map計數，排序輸出。
1 #include<bits/stdc++.h> 2 #define QAQ 0 3 #define miaojie 4 #ifdef miaojie 5 #define dbg(args...) do {cout << #args << " : "; err(args);} while (0) 6 void err() {std::cout << std::endl;} 7 template<typename T, typename...Args> 8 void err(T a, Args...args){std::cout << a << ' '; err(args...);} 9 #else 10 #define dbg(...) 11 #endif 12 13 using namespace std; 14 15 struct node{ 16 string ss; 17 int r,r2,r3; 18 bool operator <(const node a) const{ 19 if(r==a.r){ 20 if(r2==a.r2){ 21 if(r3==a.r3){ 22 return ss<a.ss; 23 } 24 return r3>a.r3; 25 } 26 return r2>a.r2; 27 } 28 return r>a.r; 29 } 30 }; 31 32 int n,a,b; 33 char c; 34 string s1,s2,s3; 35 map<string,int>m,m2,m3; 36 set<string>s; 37 38 int main(){ 39 while(cin>>n){ 40 m.clear(); m2.clear(); m3.clear(); s.clear(); 41 for(int i=1;i<=n*(n-1);i++){ 42 cin>>s1>>s2>>s3; 43 cin>>a>>c>>b; 44 s.insert(s1); s.insert(s3); 45 m2[s1]=m2[s1]+a-b; 46 m2[s3]=m2[s3]+b-a; 47 m3[s1]+=a; 48 m3[s3]+=b; 49 if(a>b){ 50 m[s1]+=3; 51 } 52 else if(a==b){ 53 m[s1]++; 54 m[s3]++; 55 } 56 else{ 57 m[s3]+=3; 58 } 59 } 60 61 node ans[n]; 62 int p=0; 63 for(auto x:s){ 64 ans[p].r=m[x]; 65 ans[p].r2=m2[x]; 66 ans[p].r3=m3[x]; 67 ans[p].ss=x; 68 p++; 69 } 70 sort(ans,ans+p); 71 for(int i=0;i<p;i++){ 72 cout<<ans[i].ss<<" "<<ans[i].r<<endl; 73 } 74 cout<<'\n'; 75 } 76 return QAQ; 77 }
hdu1226
題目給了10s...從高位到低位bfs，這樣保證一定從小到大，直到搜到模n為0的。模n相同的數等價，所以前面出現過的余數后面再有就不必入隊了。n=0特判。
1 #include<bits/stdc++.h> 2 #define QAQ 0 3 #define miaojie1 4 #ifdef miaojie 5 #define dbg(args...) do {cout << #args << " : "; err(args);} while (0) 6 void err() {std::cout << std::endl;} 7 template<typename T, typename...Args> 8 void err(T a, Args...args){std::cout << a << ' '; err(args...);} 9 #else 10 #define dbg(...) 11 #endif 12 13 using namespace std; 14 15 int T,n,c,m; 16 char ch[5]; 17 bool f[20],r[5005]; 18 19 struct node{ 20 int v[505],l,rr; 21 }; 22 23 void bfs(){ 24 queue<node>q; 25 for(int i=1;i<=15;i++){ 26 if(f[i]){ 27 node t; 28 t.v[0]=i; 29 t.rr=i%n; 30 if(r[t.rr]) continue; 31 r[t.rr]=1; 32 t.l=1; 33 if(!t.rr){ 34 if(i>=10){ 35 printf("%c",i+55); 36 } 37 else printf("%d",i); 38 return; 39 } 40 q.push(t); 41 } 42 } 43 while(!q.empty()){ 44 node t=q.front(); 45 q.pop(); 46 for(int i=0;i<=15;i++){ 47 if(f[i]){ 48 node tt=t; 49 tt.rr=(tt.rr*c+i)%n; 50 tt.v[tt.l++]=i; 51 if(!tt.rr){ 52 for(int j=0;j<tt.l;j++){ 53 if(tt.v[j]>=10){ 54 printf("%c",tt.v[j]+55); 55 } 56 else printf("%d",tt.v[j]); 57 } 58 dbg(tt.l); 59 return; 60 } 61 if(!r[tt.rr] && tt.l<=499){ 62 r[tt.rr]=1; 63 q.push(tt); 64 } 65 } 66 } 67 } 68 printf("give me the bomb please"); 69 } 70 71 int main(){ 72 scanf("%d",&T); 73 while(T--){ 74 scanf("%d%d%d",&n,&c,&m); 75 memset(f,0,sizeof(f)); 76 memset(r,0,sizeof(r)); 77 78 for(int i=1;i<=m;i++){ 79 scanf("%s",ch); 80 if(ch[0]>='A'){ 81 f[ch[0]-55]=1; 82 } 83 else f[ch[0]-48]=1; 84 } 85 if(n==0){ 86 if(f[0]){ 87 printf("0\n"); 88 } 89 else printf("give me the bomb please\n"); 90 continue; 91 } 92 bfs(); 93 printf("\n"); 94 } 95 return QAQ; 96 }
hdu 1256
根據下面內圈一定是正方形推一下各個長度。題目表達的意思是豎線寬為n/6+1....讀題讀錯了。
1 #include<bits/stdc++.h> 2 #define QAQ 0 3 4 using namespace std; 5 6 int T,n,a,b,d; 7 char c,s[4]; 8 9 void f(){ 10 for(int i=1;i<=d;i++){ 11 printf(" "); 12 } 13 for(int i=1;i<=b;i++){ 14 printf("%c",c); 15 } 16 printf("\n"); 17 } 18 19 void f2(int x){ 20 for(int i=1;i<=x;i++){ 21 for(int j=1;j<=d;j++){ 22 printf("%c",c); 23 } 24 for(int j=1;j<=b;j++){ 25 printf(" "); 26 } 27 for(int j=1;j<=d;j++){ 28 printf("%c",c); 29 } 30 printf("\n"); 31 } 32 } 33 34 int main(){ 35 scanf("%d",&T); 36 while(T--){ 37 scanf("%s%d",s,&n); 38 c=s[0]; 39 a=(n-3)/2,b=(n-2)/2,d=n/6+1; 40 f(); f2(a); 41 f(); f2(b); 42 f(); 43 if(T) printf("\n"); 44 } 45 return QAQ; 46 }
hdu 1257
轉化為求最長上升子序列。默寫板子...
1 #include<bits/stdc++.h> 2 #define QAQ 0 3 4 using namespace std; 5 const int maxn=1e6+10; 6 7 int n,a[maxn],dp[maxn]; 8 9 int main(){ 10 while(scanf("%d",&n)!=EOF){ 11 int ans=-1; 12 for(int i=1;i<=n;i++){ 13 scanf("%d",&a[i]); 14 dp[i]=1; 15 } 16 for(int i=1;i<=n;i++){ 17 for(int j=1;j<i;j++){ 18 if(a[i]>a[j]){ 19 dp[i]=max(dp[j]+1,dp[i]); 20 } 21 } 22 } 23 for(int i=1;i<=n;i++){ 24 ans=max(ans,dp[i]); 25 } 26 printf("%d\n",ans); 27 } 28 return QAQ; 29 }
hdu 1258
又是快樂dfs，題目已經排好序了~
注意每次搜的時候去重，比如搜到50的時候，下一個和再下一個都是25，只能搜1次25。也就是同一層不能搜重復數，不然最后會有相同的瘋狂刷屏.....
1 #include<bits/stdc++.h> 2 #define QAQ 0 3 4 using namespace std; 5 6 int t,n,cnt,sum,a[1000]; 7 vector<int>v; 8 9 void dfs(int x){ 10 if(sum>t) return; 11 if(sum==t){ 12 cnt++; 13 for(int i=0;i<v.size();i++){ 14 printf("%d",v[i]); 15 if(i!=v.size()-1) printf("+"); 16 else printf("\n"); 17 } 18 } 19 20 int pre=-1; 21 for(int i=x+1;i<=n;i++){ 22 if(a[i]==pre) continue; 23 pre=a[i]; 24 sum+=a[i]; 25 v.push_back(a[i]); 26 dfs(i); 27 sum-=a[i]; 28 v.pop_back(); 29 } 30 } 31 32 int main(){ 33 while(1){ 34 scanf("%d%d",&t,&n); 35 if(t==0 && n==0) break; 36 cnt=0; 37 for(int i=1;i<=n;i++){ 38 scanf("%d",&a[i]); 39 } 40 printf("Sums of %d:\n",t); 41 dfs(0); 42 if(!cnt) printf("NONE\n"); 43 } 44 return QAQ; 45 }
hdu 1259
按題意模擬。
1 #include<bits/stdc++.h> 2 #define QAQ 0 3 4 using namespace std; 5 6 int n,m,x,y; 7 8 int main(){ 9 scanf("%d",&n); 10 while(n--){ 11 string s="$ZJUTACM"; 12 scanf("%d",&m); 13 for(int i=1;i<=m;i++){ 14 scanf("%d%d",&x,&y); 15 swap(s[x],s[y]); 16 } 17 for(int i=1;i<=7;i++){ 18 if(s[i]=='J'){ 19 printf("%d\n",i); 20 break; 21 } 22 } 23 } 24 return QAQ; 25 }
hdu 1260
dp[i]表示讓i個人買完票的最短時間。第i個人買票要么單獨買，要么和前面的人合買。
狀態轉移方程dp[i]=min(dp[i-1]+s[i],dp[i-2]+d[i-1])。
1 #include<bits/stdc++.h> 2 #define QAQ 0 3 4 using namespace std; 5 6 int N,k,s[2019],d[2019],dp[2019]; 7 8 int main(){ 9 scanf("%d",&N); 10 while(N--){ 11 bool f=0; 12 scanf("%d",&k); 13 memset(dp,0,sizeof(dp)); 14 for(int i=1;i<=k;i++){ 15 scanf("%d",&s[i]); 16 } 17 for(int i=1;i<=k-1;i++){ 18 scanf("%d",&d[i]); 19 } 20 21 dp[1]=s[1]; 22 for(int i=2;i<=k;i++){ 23 dp[i]=min(dp[i-1]+s[i],dp[i-2]+d[i-1]); 24 } 25 int a=dp[k]/3600+8,b=dp[k]/60%60,c=dp[k]%60; 26 if(a>=12){ 27 a-=12; 28 f=1; 29 } 30 printf("%02d:%02d:%02d ",a,b,c); 31 if(f) printf("pm\n"); 32 else printf("am\n"); 33 } 34 return QAQ; 35 }
hdu 1261
組合數公式 sigma(a)! / sigma(a!) ，爆ll了，用大數。
1 #include<bits/stdc++.h> 2 #define QAQ 0 3 4 using namespace std; 5 6 const int base = 1000000000; 7 const int base_digits = 9; 8 9 struct Bigint { 10 vector<int> z; 11 int sign; 12 13 Bigint() : 14 sign(1) { 15 } 16 17 Bigint(long long v) { 18 *this = v; 19 } 20 21 Bigint(const string &s) { 22 read(s); 23 } 24 25 void operator=(const Bigint &v) { 26 sign = v.sign; 27 z = v.z; 28 } 29 30 void operator=(long long v) { 31 sign = 1; 32 if (v < 0) 33 sign = -1, v = -v; 34 z.clear(); 35 for (; v > 0; v = v / base) 36 z.push_back(v % base); 37 } 38 39 Bigint operator+(const Bigint &v) const { 40 if (sign == v.sign) { 41 Bigint res = v; 42 43 for (int i = 0, carry = 0; i < (int) max(z.size(), v.z.size()) || carry; ++i) { 44 if (i == (int) res.z.size()) 45 res.z.push_back(0); 46 res.z[i] += carry + (i < (int) z.size() ? z[i] : 0); 47 carry = res.z[i] >= base; 48 if (carry) 49 res.z[i] -= base; 50 } 51 return res; 52 } 53 return *this - (-v); 54 } 55 56 Bigint operator-(const Bigint &v) const { 57 if (sign == v.sign) { 58 if (abs() >= v.abs()) { 59 Bigint res = *this; 60 for (int i = 0, carry = 0; i < (int) v.z.size() || carry; ++i) { 61 res.z[i] -= carry + (i < (int) v.z.size() ? v.z[i] : 0); 62 carry = res.z[i] < 0; 63 if (carry) 64 res.z[i] += base; 65 } 66 res.trim(); 67 return res; 68 } 69 return -(v - *this); 70 } 71 return *this + (-v); 72 } 73 74 void operator*=(int v) { 75 if (v < 0) 76 sign = -sign, v = -v; 77 for (int i = 0, carry = 0; i < (int) z.size() || carry; ++i) { 78 if (i == (int) z.size()) 79 z.push_back(0); 80 long long cur = z[i] * (long long) v + carry; 81 carry = (int) (cur / base); 82 z[i] = (int) (cur % base); 83 //asm("divl %%ecx" : "=a"(carry), "=d"(a[i]) : "A"(cur), "c"(base)); 84 } 85 trim(); 86 } 87 88 Bigint operator*(int v) const { 89 Bigint res = *this; 90 res *= v; 91 return res; 92 } 93 94 friend pair<Bigint, Bigint> divmod(const Bigint &a1, const Bigint &b1) { 95 int norm = base / (b1.z.back() + 1); 96 Bigint a = a1.abs() * norm; 97 Bigint b = b1.abs() * norm; 98 Bigint q, r; 99 q.z.resize(a.z.size()); 100 101 for (int i = a.z.size() - 1; i >= 0; i--) { 102 r *= base; 103 r += a.z[i]; 104 int s1 = b.z.size() < r.z.size() ? r.z[b.z.size()] : 0; 105 int s2 = b.z.size() - 1 < r.z.size() ? r.z[b.z.size() - 1] : 0; 106 int d = ((long long) s1 * base + s2) / b.z.back(); 107 r -= b * d; 108 while (r < 0) 109 r += b, --d; 110 q.z[i] = d; 111 } 112 113 q.sign = a1.sign * b1.sign; 114 r.sign = a1.sign; 115 q.trim(); 116 r.trim(); 117 return make_pair(q, r / norm); 118 } 119 120 friend Bigint sqrt(const Bigint &a1) { 121 Bigint a = a1; 122 while (a.z.empty() || a.z.size() % 2 == 1) 123 a.z.push_back(0); 124 125 int n = a.z.size(); 126 127 int firstDigit = (int) sqrt((double) a.z[n - 1] * base + a.z[n - 2]); 128 int norm = base / (firstDigit + 1); 129 a *= norm; 130 a *= norm; 131 while (a.z.empty() || a.z.size() % 2 == 1) 132 a.z.push_back(0); 133 134 Bigint r = (long long) a.z[n - 1] * base + a.z[n - 2]; 135 firstDigit = (int) sqrt((double) a.z[n - 1] * base + a.z[n - 2]); 136 int q = firstDigit; 137 Bigint res; 138 139 for(int j = n / 2 - 1; j >= 0; j--) { 140 for(; ; --q) { 141 Bigint r1 = (r - (res * 2 * base + q) * q) * base * base + (j > 0 ? (long long) a.z[2 * j - 1] * base + a.z[2 * j - 2] : 0); 142 if (r1 >= 0) { 143 r = r1; 144 break; 145 } 146 } 147 res *= base; 148 res += q; 149 150 if (j > 0) { 151 int d1 = res.z.size() + 2 < r.z.size() ? r.z[res.z.size() + 2] : 0; 152 int d2 = res.z.size() + 1 < r.z.size() ? r.z[res.z.size() + 1] : 0; 153 int d3 = res.z.size() < r.z.size() ? r.z[res.z.size()] : 0; 154 q = ((long long) d1 * base * base + (long long) d2 * base + d3) / (firstDigit * 2); 155 } 156 } 157 158 res.trim(); 159 return res / norm; 160 } 161 162 Bigint operator/(const Bigint &v) const { 163 return divmod(*this, v).first; 164 } 165 166 Bigint operator%(const Bigint &v) const { 167 return divmod(*this, v).second; 168 } 169 170 void operator/=(int v) { 171 if (v < 0) 172 sign = -sign, v = -v; 173 for (int i = (int) z.size() - 1, rem = 0; i >= 0; --i) { 174 long long cur = z[i] + rem * (long long) base; 175 z[i] = (int) (cur / v); 176 rem = (int) (cur % v); 177 } 178 trim(); 179 } 180 181 Bigint operator/(int v) const { 182 Bigint res = *this; 183 res /= v; 184 return res; 185 } 186 187 int operator%(int v) const { 188 if (v < 0) 189 v = -v; 190 int m = 0; 191 for (int i = z.size() - 1; i >= 0; --i) 192 m = (z[i] + m * (long long) base) % v; 193 return m * sign; 194 } 195 196 void operator+=(const Bigint &v) { 197 *this = *this + v; 198 } 199 void operator-=(const Bigint &v) { 200 *this = *this - v; 201 } 202 void operator*=(const Bigint &v) { 203 *this = *this * v; 204 } 205 void operator/=(const Bigint &v) { 206 *this = *this / v; 207 } 208 209 bool operator<(const Bigint &v) const { 210 if (sign != v.sign) 211 return sign < v.sign; 212 if (z.size() != v.z.size()) 213 return z.size() * sign < v.z.size() * v.sign; 214 for (int i = z.size() - 1; i >= 0; i--) 215 if (z[i] != v.z[i]) 216 return z[i] * sign < v.z[i] * sign; 217 return false; 218 } 219 220 bool operator>(const Bigint &v) const { 221 return v < *this; 222 } 223 bool operator<=(const Bigint &v) const { 224 return !(v < *this); 225 } 226 bool operator>=(const Bigint &v) const { 227 return !(*this < v); 228 } 229 bool operator==(const Bigint &v) const { 230 return !(*this < v) && !(v < *this); 231 } 232 bool operator!=(const Bigint &v) const { 233 return *this < v || v < *this; 234 } 235 236 void trim() { 237 while (!z.empty() && z.back() == 0) 238 z.pop_back(); 239 if (z.empty()) 240 sign = 1; 241 } 242 243 bool isZero() const { 244 return z.empty() || (z.size() == 1 && !z[0]); 245 } 246 247 Bigint operator-() const { 248 Bigint res = *this; 249 res.sign = -sign; 250 return res; 251 } 252 253 Bigint abs() const { 254 Bigint res = *this; 255 res.sign *= res.sign; 256 return res; 257 } 258 259 long long longValue() const { 260 long long res = 0; 261 for (int i = z.size() - 1; i >= 0; i--) 262 res = res * base + z[i]; 263 return res * sign; 264 } 265 266 friend Bigint gcd(const Bigint &a, const Bigint &b) { 267 return b.isZero() ? a : gcd(b, a % b); 268 } 269 friend Bigint lcm(const Bigint &a, const Bigint &b) { 270 return a / gcd(a, b) * b; 271 } 272 273 void read(const string &s) { 274 sign = 1; 275 z.clear(); 276 int pos = 0; 277 while (pos < (int) s.size() && (s[pos] == '-' || s[pos] == '+')) { 278 if (s[pos] == '-') 279 sign = -sign; 280 ++pos; 281 } 282 for (int i = s.size() - 1; i >= pos; i -= base_digits) { 283 int x = 0; 284 for (int j = max(pos, i - base_digits + 1); j <= i; j++) 285 x = x * 10 + s[j] - '0'; 286 z.push_back(x); 287 } 288 trim(); 289 } 290 291 friend istream& operator>>(istream &stream, Bigint &v) { 292 string s; 293 stream >> s; 294 v.read(s); 295 return stream; 296 } 297 298 friend ostream& operator<<(ostream &stream, const Bigint &v) { 299 if (v.sign == -1) 300 stream << '-'; 301 stream << (v.z.empty() ? 0 : v.z.back()); 302 for (int i = (int) v.z.size() - 2; i >= 0; --i) 303 stream << setw(base_digits) << setfill('0') << v.z[i]; 304 return stream; 305 } 306 307 static vector<int> convert_base(const vector<int> &a, int old_digits, int new_digits) { 308 vector<long long> p(max(old_digits, new_digits) + 1); 309 p[0] = 1; 310 for (int i = 1; i < (int) p.size(); i++) 311 p[i] = p[i - 1] * 10; 312 vector<int> res; 313 long long cur = 0; 314 int cur_digits = 0; 315 for (int i = 0; i < (int) a.size(); i++) { 316 cur += a[i] * p[cur_digits]; 317 cur_digits += old_digits; 318 while (cur_digits >= new_digits) { 319 res.push_back(int(cur % p[new_digits])); 320 cur /= p[new_digits]; 321 cur_digits -= new_digits; 322 } 323 } 324 res.push_back((int) cur); 325 while (!res.empty() && res.back() == 0) 326 res.pop_back(); 327 return res; 328 } 329 330 typedef vector<long long> vll; 331 332 static vll karatsubaMultiply(const vll &a, const vll &b) { 333 int n = a.size(); 334 vll res(n + n); 335 if (n <= 32) { 336 for (int i = 0; i < n; i++) 337 for (int j = 0; j < n; j++) 338 res[i + j] += a[i] * b[j]; 339 return res; 340 } 341 342 int k = n >> 1; 343 vll a1(a.begin(), a.begin() + k); 344 vll a2(a.begin() + k, a.end()); 345 vll b1(b.begin(), b.begin() + k); 346 vll b2(b.begin() + k, b.end()); 347 348 vll a1b1 = karatsubaMultiply(a1, b1); 349 vll a2b2 = karatsubaMultiply(a2, b2); 350 351 for (int i = 0; i < k; i++) 352 a2[i] += a1[i]; 353 for (int i = 0; i < k; i++) 354 b2[i] += b1[i]; 355 356 vll r = karatsubaMultiply(a2, b2); 357 for (int i = 0; i < (int) a1b1.size(); i++) 358 r[i] -= a1b1[i]; 359 for (int i = 0; i < (int) a2b2.size(); i++) 360 r[i] -= a2b2[i]; 361 362 for (int i = 0; i < (int) r.size(); i++) 363 res[i + k] += r[i]; 364 for (int i = 0; i < (int) a1b1.size(); i++) 365 res[i] += a1b1[i]; 366 for (int i = 0; i < (int) a2b2.size(); i++) 367 res[i + n] += a2b2[i]; 368 return res; 369 } 370 371 Bigint operator*(const Bigint &v) const { 372 vector<int> a6 = convert_base(this->z, base_digits, 6); 373 vector<int> b6 = convert_base(v.z, base_digits, 6); 374 vll a(a6.begin(), a6.end()); 375 vll b(b6.begin(), b6.end()); 376 while (a.size() < b.size()) 377 a.push_back(0); 378 while (b.size() < a.size()) 379 b.push_back(0); 380 while (a.size() & (a.size() - 1)) 381 a.push_back(0), b.push_back(0); 382 vll c = karatsubaMultiply(a, b); 383 Bigint res; 384 res.sign = sign * v.sign; 385 for (int i = 0, carry = 0; i < (int) c.size(); i++) { 386 long long cur = c[i] + carry; 387 res.z.push_back((int) (cur % 1000000)); 388 carry = (int) (cur / 1000000); 389 } 390 res.z = convert_base(res.z, 6, base_digits); 391 res.trim(); 392 return res; 393 } 394 }; 395 396 int n; 397 398 int main(){ 399 while(scanf("%d",&n)){ 400 if(n==0) break; 401 Bigint ans=1; 402 int a[30],cnt=0; 403 for(int i=1;i<=n;i++){ 404 cin>>a[i]; 405 cnt+=a[i]; 406 } 407 for(int i=1;i<=cnt;i++){ 408 ans*=i; 409 } 410 for(int i=1;i<=n;i++){ 411 for(int j=1;j<=a[i];j++){ 412 ans/=j; 413 } 414 } 415 cout<<ans<<endl; 416 } 417 return QAQ; 418 }
hdu 1262
線性篩預處理一下素數，從m/2開始搜，第一個搜到的一定是最近的。
1 #include<bits/stdc++.h> 2 #define QAQ 0 3 4 using namespace std; 5 const int maxn=100000; 6 int prime[maxn],check[maxn]; 7 int m,n; 8 9 void xxs(){ 10 int tot=0; 11 memset(check,0,sizeof(check)); 12 check[1]=1; 13 for(int i=2;i<maxn;i++){ 14 if(!check[i]) prime[tot++]=i; 15 for(int j=0;j<tot;j++){ 16 if(i*prime[j]>maxn) break; 17 check[i*prime[j]]=1; 18 if(i%prime[j]==0) break; 19 } 20 } 21 } 22 23 int main(){ 24 xxs(); 25 while(scanf("%d",&m)!=EOF){ 26 n=m/2; 27 while(1){ 28 if(!check[n] && !check[m-n]){ 29 printf("%d %d\n",n,m-n); 30 break; 31 } 32 n--; 33 } 34 } 35 return QAQ; 36 }
hdu 1263
二維map計數。
1 #include<bits/stdc++.h> 2 #define QAQ 0 3 4 using namespace std; 5 6 int N,M,a; 7 map<string,map<string,int> >m; 8 string s1,s2; 9 10 int main(){ 11 cin>>N; 12 while(N--){ 13 m.clear(); 14 cin>>M; 15 for(int i=1;i<=M;i++){ 16 cin>>s1>>s2>>a; 17 m[s2][s1]+=a; 18 } 19 for(auto x:m){ 20 cout<<x.first<<'\n'; 21 for(auto y:x.second){ 22 cout<<" |----"<<y.first<<'('<<y.second<<')'<<'\n'; 23 } 24 } 25 if(N) cout<<'\n'; 26 } 27 return QAQ; 28 }
hdu 1264
注意數據范圍0到100...直接暴力掃。確定矩形位置的兩個點有兩種情況，一開始沒swap掉坑里了。
1 #include<bits/stdc++.h> 2 #define QAQ 0 3 4 using namespace std; 5 6 int a1,b1,a2,b2; 7 bool m[102][102]; 8 9 int main(){ 10 while(1){ 11 memset(m,0,sizeof(m)); 12 int cnt=0; 13 while(1){ 14 scanf("%d%d%d%d",&a1,&b1,&a2,&b2); 15 if(a1==-1 || a1==-2) break; 16 if(a1>a2) swap(a1,a2); 17 if(b1>b2) swap(b1,b2); 18 19 for(int i=b1;i<b2;i++){ 20 for(int j=a1;j<a2;j++){ 21 if(!m[i][j]){ 22 cnt++; 23 m[i][j]=1; 24 } 25 } 26 } 27 } 28 printf("%d\n",cnt); 29 if(a1==-2) break; 30 } 31 return QAQ; 32 }
hdu 1265
十六進制表示浮點數...挺奇特的...完全沒想到這個思路。直接把計算機存放的二進制化成十六進制。。。
1 #include<bits/stdc++.h> 2 #define QAQ 0 3 4 using namespace std; 5 6 int N; 7 float f; 8 9 int main(){ 10 scanf("%d",&N); 11 while(N--){ 12 scanf("%f",&f); 13 unsigned char *p=(unsigned char *)&f; 14 printf("%02X%02X%02X%02X\n",p[3],p[2],p[1],p[0]); 15 } 16 return QAQ; 17 }
hdu 1491
模擬題。
1 #include<bits/stdc++.h> 2 #define QAQ 0 3 4 using namespace std; 5 6 int T,a,b; 7 int m[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}; 8 9 int main(){ 10 scanf("%d",&T); 11 while(T--){ 12 scanf("%d%d",&a,&b); 13 if(a>10 || (a==10&&b>21)){ 14 printf("What a pity, it has passed!\n"); 15 } 16 else if(a==10 && b==21){ 17 printf("It's today!!\n"); 18 } 19 else{ 20 if(a==10){ 21 printf("%d\n",21-b); 22 } 23 else{ 24 int ans=21+m[a]-b; 25 for(int i=a+1;i<=9;i++){ 26 ans+=m[i]; 27 } 28 printf("%d\n",ans); 29 } 30 } 31 } 32 return QAQ; 33 }
hdu 1492
因子只有2，3，5，7。統計它們的冪次，用求因數個數的公式就好。不開ll會wa。
1 #include<bits/stdc++.h> 2 #define QAQ 0 3 4 using namespace std; 5 typedef long long ll; 6 7 ll n; 8 ll f[4]={2,3,5,7},cnt[4]; 9 10 int main(){ 11 while(1){ 12 scanf("%lld",&n); 13 if(n==0) break; 14 ll ans=1; 15 for(int i=0;i<4;i++){ 16 ll t=n; 17 cnt[i]=0; 18 while(t%f[i]==0){ 19 cnt[i]++; 20 t/=f[i]; 21 } 22 ans*=(cnt[i]+1); 23 } 24 printf("%lld\n",ans); 25 } 26 return QAQ; 27 }
hdu 1493
概率dp。本來想開二維，其實可以滾動，每次的概率以上次為基礎。每次從后往前，dp[i]=dp[i]+dp[i-k]*rp[k]，k為擲出的點數。
1 #include<bits/stdc++.h> 2 #define QAQ 0 3 #define miaojie1 4 #ifdef miaojie 5 #define dbg(args...) do {cout << #args << " : "; err(args);} while (0) 6 void err() {std::cout << std::endl;} 7 template<typename T, typename...Args> 8 void err(T a, Args...args){std::cout << a << ' '; err(args...);} 9 #else 10 #define dbg(...) 11 #endif 12 13 using namespace std; 14 15 int T; 16 double rp[10],dp[100]; 17 int a[11]={0,5,12,22,29,33,38,42,46,50,55}; 18 19 int main(){ 20 scanf("%d",&T); 21 22 while(T--){ 23 for(int i=1;i<=6;i++){ 24 scanf("%lf",&rp[i]); 25 } 26 memset(dp,0,sizeof(dp)); 27 dp[0]=1; 28 29 for(int i=1;i<=10;i++){ 30 dbg(dp[12]); 31 for(int j=60;j>0;j--){ 32 dp[j]=0; 33 for(int k=1;k<=6;k++){ 34 if(j<k) break; 35 dp[j]=dp[j]+rp[k]*dp[j-k]; 36 } 37 } 38 } 39 for(int i=1;i<=10;i++){ 40 printf("%d: %.1lf%%\n",a[i],100.0*dp[a[i]]); 41 } 42 if(T) printf("\n"); 43 } 44 return QAQ; 45 }
hdu 1494
還是dp。把n圈展開成線，能量看成0-14的整數，dp[i][j]表示到第i段時有j格能量所用的最短時間，順序dp。注意j=0一定是剛用過加速卡，j=10可能是普通行駛或爆卡，j>10一定是普通行駛...分類挺麻煩的，狀態方程分四個。感覺還可以簡化下。
1 #include<bits/stdc++.h> 2 #define QAQ 0 3 #define miaojie1 4 #ifdef miaojie 5 #define dbg(args...) do {cout << #args << " : "; err(args);} while (0) 6 void err() {std::cout << std::endl;} 7 template<typename T, typename...Args> 8 void err(T a, Args...args){std::cout << a << ' '; err(args...);} 9 #else 10 #define dbg(...) 11 #endif 12 13 using namespace std; 14 typedef long long ll; 15 16 int l,n; 17 ll a[10020],b[10020],dp[10020][15]; 18 19 int main(){ 20 while(scanf("%d",&l)!=EOF){ 21 scanf("%d",&n); 22 int sum=n*l; 23 for(int i=1;i<=l;i++){ 24 scanf("%d",&a[i]); 25 for(int j=1;j<=n-1;j++){ 26 a[i+j*l]=a[i]; 27 } 28 } 29 for(int i=1;i<=l;i++){ 30 scanf("%d",&b[i]); 31 for(int j=1;j<=n-1;j++){ 32 b[i+j*l]=b[i]; 33 } 34 } 35 for(int i=0;i<=sum;i++){ 36 for(int j=0;j<15;j++){ 37 dp[i][j]=1ll<<30; 38 } 39 } 40 dp[0][0]=0; 41 42 for(int i=1;i<=sum;i++){ 43 for(int j=0;j<15;j++){ 44 if(j==0) { 45 dp[i][j]=dp[i-1][5]+b[i]; 46 } 47 else if(j==10){ 48 dp[i][j]=min( dp[i-1][14]+a[i] , dp[i-1][9]+a[i] ); 49 } 50 else if(j>10){ 51 dp[i][j]=dp[i-1][j-1]+a[i]; 52 } 53 else dp[i][j]=min( dp[i-1][j-1]+a[i] , dp[i-1][j+5]+b[i] ); 54 } 55 } 56 ll ans=1ll<<62; 57 for(int i=0;i<15;i++){ 58 ans=min(ans,dp[sum][i]); 59 dbg(dp[sum][i]); 60 } 61 printf("%lld\n",ans); 62 } 63 return QAQ; 64 }
hdu 1495
bfs，模擬每步的六種操作即可。
1 #include<bits/stdc++.h> 2 #define miaojie1 3 #ifdef miaojie 4 #define dbg(args...) do {cout << #args << " : "; err(args);} while (0) 5 void err() {std::cout << std::endl;} 6 template<typename T, typename...Args> 7 void err(T a, Args...args){std::cout << a << ' '; err(args...);} 8 #else 9 #define dbg(...) 10 #endif 11 12 using namespace std; 13 14 int S,s,n,m; 15 bool vis[105][105][105]; 16 17 struct node{ 18 int a,b,c,r; 19 node(){} 20 node(int _a,int _b,int _c,int _r) : a(_a),b(_b),c(_c),r(_r){} 21 }; 22 23 void bfs(){ 24 memset(vis,0,sizeof(vis)); 25 if(S%2){ 26 printf("NO\n"); 27 return; 28 } 29 else s=S/2; 30 queue<node>q; 31 node st(S,0,0,0); 32 q.push(st); 33 vis[s][0][0]=1; 34 35 while(!q.empty()){ 36 node t=q.front(); 37 dbg(t.a,t.b,t.c,t.r); 38 if( (t.a==s&&t.b==s) || (t.a==s&&t.c==s) || (t.b==s&&t.c==s)){ 39 printf("%d\n",t.r); 40 return; 41 } 42 q.pop(); 43 if(t.a>0){ 44 node p=t; 45 if(p.a>=n-p.b){ 46 p.a=p.a-(n-p.b); 47 p.b=n; 48 p.r++; 49 } 50 else{ 51 p.b+=p.a; 52 p.a=0; 53 p.r++; 54 } 55 if(!vis[p.a][p.b][p.c]) q.push(p),vis[p.a][p.b][p.c]=1; 56 57 p=t; 58 if(p.a>=m-p.c){ 59 p.a=p.a-(m-p.c); 60 p.c=m; 61 p.r++; 62 } 63 else{ 64 p.c+=p.a; 65 p.a=0; 66 p.r++; 67 } 68 if(!vis[p.a][p.b][p.c]) q.push(p),vis[p.a][p.b][p.c]=1; 69 70 p=t; 71 if(p.b>=S-p.a){ 72 p.b=p.b-(S-p.a); 73 p.a=s; 74 p.r++; 75 } 76 else{ 77 p.a+=p.b; 78 p.b=0; 79 p.r++; 80 } 81 if(!vis[p.a][p.b][p.c]) q.push(p),vis[p.a][p.b][p.c]=1; 82 83 p=t; 84 if(p.b>=m-p.c){ 85 p.b=p.b-(m-p.c); 86 p.c=m; 87 p.r++; 88 } 89 else{ 90 p.c+=p.b; 91 p.b=0; 92 p.r++; 93 } 94 if(!vis[p.a][p.b][p.c]) q.push(p),vis[p.a][p.b][p.c]=1; 95 96 p=t; 97 if(p.c>=S-p.a){ 98 p.c=p.c-(S-p.a); 99 p.a=s; 100 p.r++; 101 } 102 else{ 103 p.a+=p.c; 104 p.c=0; 105 p.r++; 106 } 107 if(!vis[p.a][p.b][p.c]) q.push(p),vis[p.a][p.b][p.c]=1; 108 109 p=t; 110 if(p.c>=n-p.b){ 111 p.c=p.c-(n-p.b); 112 p.b=n; 113 p.r++; 114 } 115 else{ 116 p.b+=p.c; 117 p.c=0; 118 p.r++; 119 } 120 if(!vis[p.a][p.b][p.c]) q.push(p),vis[p.a][p.b][p.c]=1; 121 } 122 } 123 printf("NO\n"); 124 } 125 126 int main(){ 127 while(1){ 128 scanf("%d%d%d",&S,&n,&m); 129 if(S==0 && n==0 && m==0) break; 130 bfs(); 131 } 132 return 0; 133 }
hdu 1496
四個for顯然超了，用折半法降時間復雜度，結果還是T，加了特判才過。由於有平方，只討論正號省時間，最后乘16就行。
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 typedef long long ll; 5 6 int a,b,c,d; 7 map<int,int>mp; 8 9 int main(){ 10 while(scanf("%d%d%d%d",&a,&b,&c,&d)!=EOF){ 11 if((a>0 && b>0 && c>0 && d>0) || (a<0 && b<0 && c<0 && d<0)){ 12 printf("0\n"); 13 continue; 14 } 15 ll ans=0; 16 mp.clear(); 17 for(int i=1;i<=100;i++){ 18 for(int j=1;j<=100;j++){ 19 mp[a*i*i+b*j*j]++; 20 } 21 } 22 for(int i=1;i<=100;i++){ 23 for(int j=1;j<=100;j++){ 24 ans+=mp[-(c*i*i+d*j*j)]; 25 } 26 } 27 ans*=16; 28 printf("%lld\n",ans); 29 } 30 return 0; 31 }
hdu 1497
怎么又是大模擬，8想做！不看它了！
hdu 1498
行對列匹配跑一遍二分圖的最小覆蓋,balabala（明天補上）
第二天：好像也補不上啥啊。。多一個匹配就多一個既不同行也不同列的氣球，超過k個就不行了。
1 #include<bits/stdc++.h> 2 #define miaojie1 3 #ifdef miaojie 4 #define dbg(args...) do {cout << #args << " : "; err(args);} while (0) 5 void err() {std::cout << std::endl;} 6 template<typename T, typename...Args> 7 void err(T a, Args...args){std::cout << a << ' '; err(args...);} 8 #else 9 #define dbg(...) 10 #endif 11 using namespace std; 12 13 int n,k,a[105][105],edge[105][105],cx[105],cy[105],ans[105],pp; 14 bool vis[105]; 15 set<int>se; 16 17 int line(int u){ 18 for(int v=1;v<=n;v++){ 19 if(edge[u][v] && !vis[v]){ 20 vis[v]=1; 21 if(cy[v]==-1 || line(cy[v])){ 22 cx[u]=v; 23 cy[v]=u; 24 return 1; 25 } 26 } 27 } 28 return 0; 29 } 30 31 int mincover(){ 32 int ret=0; 33 memset(cx,-1,sizeof(cx)); 34 memset(cy,-1,sizeof(cy)); 35 for(int i=1;i<=n;i++){ 36 if(cx[i]==-1){ 37 memset(vis,0,sizeof(vis)); 38 ret+=line(i); 39 } 40 } 41 return ret; 42 } 43 44 int main(){ 45 while(1){ 46 scanf("%d%d",&n,&k); 47 se.clear(); pp=0; 48 if(n==0 && k==0) break; 49 for(int i=1;i<=n;i++){ 50 for(int j=1;j<=n;j++){ 51 scanf("%d",&a[i][j]); 52 se.insert(a[i][j]); 53 } 54 } 55 for(auto x:se){ 56 dbg(x); 57 memset(edge,0,sizeof(edge)); 58 for(int i=1;i<=n;i++){ 59 for(int j=1;j<=n;j++){ 60 if(a[i][j]==x){ 61 edge[i][j]=1; 62 } 63 } 64 } 65 if(mincover()>k) ans[pp++]=x; 66 } 67 dbg(pp); 68 if(pp==0){ 69 printf("-1\n"); 70 } 71 else{ 72 for(int i=0;i<pp;i++){ 73 printf("%d",ans[i]); 74 if(i!=pp-1) printf(" "); 75 else printf("\n"); 76 } 77 } 78 } 79 return 0; 80 }
hdu 1728
一開始傳統bfs，出隊一個就搜四周，記錄朝向來表示是否轉彎，mle了。
其實這題應該換個想法，答案只和拐彎數有關，所以搜的時候可以不再前進一個點，改為一直走到某方向的盡頭，路上的點還是判vis按順序入隊。這樣每個點搜出一個十字形狀，並且保證了第一次搜到時拐彎數最小。
定義了y1，沒敢用萬能頭文件。
1 #include<iostream> 2 #include<cstdlib> 3 #include<cstdio> 4 #include<algorithm> 5 #include<queue> 6 #include<string.h> 7 using namespace std; 8 9 int T,m,n; 10 char a[105][105]; 11 bool vis[105][105]; 12 int k,x1,x2,y1,y2; 13 int tx[5]={0,-1,0,1,0},ty[5]={0,0,1,0,-1}; 14 15 struct node{ 16 int x,y,r; 17 node(){} 18 node(int _x,int _y,int _r) : x(_x),y(_y),r(_r){} 19 }; 20 21 void bfs(){ 22 queue<node>q; 23 vis[x1][y1]=1; 24 // node st1(x1,y1,-1,1),st2(x1,y1,-1,2),st3(x1,y1,-1,3),st4(x1,y1,-1,4); 25 // q.push(st1); q.push(st2); q.push(st3); q.push(st4); 26 node st(x1,y1,-1); 27 q.push(st); 28 while(!q.empty()){ 29 node t=q.front(); 30 // cout<<"!!!"<<t.x<<" "<<t.y<<" "<<t.r<<" "<<endl; 31 q.pop(); 32 if(t.r>k) continue; 33 if(t.x==x2 && t.y==y2){ 34 // cout<<"~~~"<<t.r<<" "<<t.s<<endl; 35 printf("yes\n"); 36 return; 37 } 38 for(int i=1;i<=4;i++){ 39 node p=t; 40 p.x+=tx[i]; 41 p.y+=ty[i]; 42 p.r++; 43 while(!(p.x<1||p.x>m||p.y<1||p.y>n||p.r>k||a[p.x][p.y]=='*')){ 44 if(!vis[p.x][p.y]){ 45 vis[p.x][p.y]=1; 46 q.push(p); 47 } 48 p.x+=tx[i]; 49 p.y+=ty[i]; 50 } 51 } 52 } 53 printf("no\n"); 54 } 55 56 int main(){ 57 scanf("%d",&T); 58 while(T--){ 59 scanf("%d%d",&m,&n); 60 memset(vis,0,sizeof(vis)); 61 for(int i=1;i<=m;i++){ 62 for(int j=1;j<=n;j++){ 63 cin>>a[i][j]; 64 } 65 } 66 scanf("%d%d%d%d%d",&k,&y1,&x1,&y2,&x2); 67 bfs(); 68 } 69 return 0; 70 }
hdu 1729
sg函數太生疏了.....
s為容量，找到臨界點t+t*t<s且(t+1)+(t+1)^2>=s，可知當盒子里大於t個時必勝，返回容量減現有個數（類比取石子），等於t時必敗。小於t時無法判斷，但s和t都是必敗點，返回（t，c）等價。
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 int n,cnt,s,c; 6 7 int SG(int s,int c){ 8 int t=sqrt(s); 9 while(t+t*t>=s){ 10 t--; 11 } 12 if(c>t) return s-c; 13 else return SG(t,c); 14 } 15 16 int main(){ 17 while(1){ 18 scanf("%d",&n); 19 if(n==0) break; 20 int ans=0; 21 for(int i=1;i<=n;i++){ 22 scanf("%d%d",&s,&c); 23 ans^=SG(s,c); 24 } 25 printf("Case %d:\n",++cnt); 26 if(ans) printf("Yes\n"); 27 else printf("No\n"); 28 } 29 return 0; 30 }
hdu 1730
這題友好多了，每行兩個棋子的距離-1其實就類比每堆石子數，nim一下就好。
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 int n,m,a,b; 6 7 int main(){ 8 while(scanf("%d%d",&n,&m)!=EOF){ 9 int ans=0; 10 for(int i=1;i<=n;i++){ 11 scanf("%d%d",&a,&b); 12 ans=ans^(abs(a-b)-1); 13 } 14 if(!ans){ 15 printf("BAD LUCK!\n"); 16 } 17 else printf("I WIN!\n"); 18 } 19 return 0; 20 }
hdu 1732
bfs，記錄人和三個箱子的位置。走一步——是否有障礙或箱子——若有箱子，箱子后面是否有障礙或其他箱子。
因為把=寫成==爆內存了，找了一個小時，強行拉底ac率，很煩。
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 struct node{ 6 int x[4],y[4],r; 7 }st; 8 9 int n,m; 10 int tx[4]={-1,0,1,0},ty[4]={0,1,0,-1},endx[4],endy[4]; 11 char a[8][8]; 12 bool vis[8][8][8][8][8][8][8][8]; 13 queue<node>q; 14 void bfs(){ 15 memset(vis,0,sizeof(vis)); 16 while(!q.empty()) q.pop(); 17 q.push(st); 18 while(!q.empty()){ 19 node t=q.front(); 20 q.pop(); 21 bool check=1; 22 for(int i=1;i<=3;i++){ 23 bool ck=0; 24 for(int j=1;j<=3;j++){ 25 if(t.x[i]==endx[j] && t.y[i]==endy[j]) ck=1; 26 } 27 if(!ck){ 28 check=0; 29 break; 30 } 31 } 32 if(check){ 33 printf("%d\n",t.r); 34 return; 35 } 36 if(vis[t.x[0]][t.y[0]][t.x[1]][t.y[1]][t.x[2]][t.y[2]][t.x[3]][t.y[3]]==1) continue; 37 vis[t.x[0]][t.y[0]][t.x[1]][t.y[1]][t.x[2]][t.y[2]][t.x[3]][t.y[3]]==1; 38 for(int i=0;i<=3;i++){ 39 node p=t; 40 int f=0; 41 p.x[0]+=tx[i]; 42 p.y[0]+=ty[i]; 43 p.r++; 44 if(a[p.x[0]][p.y[0]]=='#'||p.x[0]>=n||p.x[0]<0||p.y[0]>=m||p.y[0]<0) continue; 45 for(int j=1;j<=3;j++){ 46 if(p.x[0]==p.x[j] && p.y[0]==p.y[j]){ 47 f=j; 48 break; 49 } 50 } 51 if(f){ 52 p.x[f]+=tx[i]; 53 p.y[f]+=ty[i]; 54 if(a[p.x[f]][p.y[f]]=='#'||p.x[f]>=n||p.x[f]<0||p.y[f]>=m||p.y[f]<0) continue; 55 bool ff=0; 56 for(int j=1;j<=3;j++){ 57 if(j==f) continue; 58 if(p.x[f]==p.x[j] && p.y[f]==p.y[j]){ 59 ff=1; 60 break; 61 } 62 } 63 if(!ff) q.push(p); 64 } 65 else{ 66 q.push(p); 67 } 68 } 69 } 70 printf("-1\n"); 71 } 72 73 int main(){ 74 ios_base::sync_with_stdio(false),cin.tie(NULL),cout.tie(NULL); 75 while(cin>>n>>m){ 76 int p=0,q=0; 77 for(int i=0;i<n;i++){ 78 for(int j=0;j<m;j++){ 79 cin>>a[i][j]; 80 if(a[i][j]=='X'){ 81 st.x[0]=i; 82 st.y[0]=j; 83 } 84 else if(a[i][j]=='*'){ 85 st.x[++p]=i; 86 st.y[p]=j; 87 } 88 else if(a[i][j]=='@'){ 89 endx[++q]=i; 90 endy[q]=j; 91 } 92 } 93 } 94 st.r=0; 95 bfs(); 96 } 97 return 0; 98 }
hdu 5702
排序。
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 struct node{ 6 char s[20]; 7 int v; 8 bool operator < (const node &t) const{ 9 return v>t.v; 10 } 11 }a[20]; 12 13 int T,n; 14 15 int main(){ 16 scanf("%d",&T); 17 while(T--){ 18 scanf("%d",&n); 19 for(int i=1;i<=n;i++){ 20 scanf("%s%d",a[i].s,&a[i].v); 21 } 22 sort(a+1,a+1+n); 23 for(int i=1;i<=n;i++){ 24 printf("%s",a[i].s); 25 if(i==n) printf("\n"); 26 else printf(" "); 27 } 28 } 29 return 0; 30 }
hdu 5703
列出前幾項就能發現規律。
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 int T,n; 6 7 int main(){ 8 scanf("%d",&T); 9 while(T--){ 10 scanf("%d",&n); 11 printf("1"); 12 for(int i=1;i<=n-1;i++){ 13 printf("0"); 14 } 15 printf("\n"); 16 } 17 return 0; 18 }
hdu 5704
方程列出來，解方程。
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 int T,n,sum,a[102]; 6 map<int,int>m; 7 8 int main(){ 9 scanf("%d",&T); 10 while(T--){ 11 m.clear(); sum=0; 12 scanf("%d",&n); 13 for(int i=1;i<=n-1;i++){ 14 scanf("%d",&a[i]); 15 m[a[i]]++; 16 sum+=a[i]; 17 } 18 int ans=2*sum/(3*n-2); 19 double p=1.0/(m[ans]+1.0); 20 printf("%d %.2lf\n",ans,p); 21 } 22 return 0; 23 }
hdu 5705
計算時針和分針的移動速度，從00：00：00開始枚舉判斷角度。為了避開精度問題乘了11。
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 int h,m,s,a,cas,ansh,ansm,anss; 6 7 int main(){ 8 ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL); 9 while(~scanf("%d:%d:%d%d",&h,&m,&s,&a)){ 10 printf("Case #%d: ",++cas); 11 int t=(h*3600+m*60+s)*11,tim=0,cnt=0; 12 while(1){ 13 tim+=120; 14 cnt+=11; 15 int tmp=cnt%3960; 16 if(tmp>1980) tmp=3960-tmp; 17 if(tmp==a*11&&tim>t) break; 18 } 19 tim/=11; 20 tim=tim%(12*60*60); 21 ansh=tim/3600; 22 ansm=(tim-ansh*3600)/60; 23 anss=tim-ansh*3600-ansm*60; 24 printf("%02d:%02d:%02d\n",ansh,ansm,anss); 25 } 26 return 0; 27 }
hdu 5706
dfs連通塊，搜就行了。
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 int T,n,m,ans1,ans2; 6 char a[1005][1005]; 7 char s1[10]="girl",s2[10]="cat"; 8 9 void dfs1(int i,int j,int k){ 10 if(i<1||i>n||j<1||j>m) return; 11 if(a[i][j]==s1[k]){ 12 k++; 13 if(k==4){ 14 ans1++; 15 return; 16 } 17 dfs1(i-1,j,k); 18 dfs1(i+1,j,k); 19 dfs1(i,j+1,k); 20 dfs1(i,j-1,k); 21 } 22 } 23 24 void dfs2(int i,int j,int k){ 25 if(i<1||i>n||j<1||j>m) return; 26 if(a[i][j]==s2[k]){ 27 k++; 28 if(k==3){ 29 ans2++; 30 return; 31 } 32 dfs2(i-1,j,k); 33 dfs2(i+1,j,k); 34 dfs2(i,j+1,k); 35 dfs2(i,j-1,k); 36 } 37 } 38 39 int main(){ 40 scanf("%d",&T); 41 while(T--){ 42 ans1=ans2=0; 43 scanf("%d%d",&n,&m); 44 for(int i=1;i<=n;i++){ 45 scanf("%s",a[i]+1); 46 } 47 for(int i=1;i<=n;i++){ 48 for(int j=1;j<=m;j++){ 49 if(a[i][j]=='g') dfs1(i,j,0); 50 if(a[i][j]=='c') dfs2(i,j,0); 51 } 52 } 53 printf("%d %d\n",ans1,ans2); 54 } 55 return 0; 56 }
hdu 5707
由於c中所有字母都要使用，按順序對於每個字母判斷是否能加入即可。dp兩維分別表示a和b能匹配的字母數。
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 bool dp[2002][2002]; 6 char a[2002],b[2002],c[2002]; 7 8 int main(){ 9 while(~scanf("%s%s%s",a,b,c)){ 10 int la=strlen(a),lb=strlen(b),lc=strlen(c); 11 if(lc!=la+lb){ 12 printf("No\n"); 13 continue; 14 } 15 memset(dp,0,sizeof(dp)); 16 dp[0][0]=1; 17 for(int i=0;i<=la;i++){ 18 for(int j=0;j<=lb;j++){ 19 if(c[i+j]==a[i]){ 20 dp[i+1][j]|=dp[i][j]; 21 } 22 if(c[i+j]==b[j]){ 23 dp[i][j+1]|=dp[i][j]; 24 } 25 } 26 } 27 if(dp[la][lb]) printf("Yes\n"); 28 else printf("No\n"); 29 } 30 return 0; 31 }
hdu5708
sg打表找規律.....看了半天，周期居然是2*（k+1）。k=1需要特判。
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 const int N=30; 5 6 int T,q,k,n,m; 7 int sg[50][50]; 8 9 void get_SG(){ 10 sg[1][1]=0; 11 for(int i=2;i<=N;i+=2){ 12 sg[i][1]=sg[1][i]=1; 13 sg[i+1][1]=sg[1][i+1]=0; 14 } 15 for(int i=2;i<=N;i++){ 16 for(int j=2;j<=N;j++){ 17 if(!sg[i-1][j] || !sg[i][j-1] || (i-k>0 && j-k>0 && (!sg[i-k][j-k])) ) 18 sg[i][j]=1; 19 else sg[i][j]=0; 20 } 21 } 22 } 23 24 void p1(){ 25 printf("Alice\n"); 26 } 27 28 void p2(){ 29 printf("Bob\n"); 30 } 31 32 int main(){ 33 scanf("%d",&T); 34 while(T--){ 35 scanf("%d%d",&q,&k); 36 /* memset(sg,0,sizeof(sg)); 37 get_SG(); 38 printf("!!!\n"); 39 for(int i=1;i<=N;i++){ 40 for(int j=1;j<=N;j++){ 41 printf("%d ",sg[i][j]); 42 } 43 printf("\n"); 44 } 45 printf("!!!\n\n"); */ 46 47 while(q--){ 48 scanf("%d%d",&n,&m); 49 int l=min(n,m),s=n+m; 50 if(k==1){ 51 if(l%2){ 52 if(s%2) p1(); 53 else p2(); 54 } 55 else p1(); 56 continue; 57 } 58 int t=(k+1)*2; 59 if(l%t>=1&&l%t<=k){ 60 if(s%2) p1(); 61 else p2(); 62 } 63 else if(l%(k+1)==0){ 64 p1(); 65 } 66 else{ 67 if(s%2) p2(); 68 else p1(); 69 } 70 } 71 } 72 return 0; 73 }
hdu 5710
容易發現0~4乘2后不損失值，5~9乘2后損失9。設大於5的部分長為l，得到S(n)與l的比例關系，顯然約分后數字最小。從低位開始放數字，讓低位的盡量大。
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 int ans[1000000],p,a,b,T,s,l; 6 7 int main(){ 8 scanf("%d",&T); 9 while(T--){ 10 scanf("%d%d",&a,&b); 11 p=0; 12 s=9*b; 13 l=2*b-a; 14 int t=__gcd(s,l); 15 s/=t; l/=t; 16 s=s-l*5; 17 if(l<0||s<0){ 18 puts("0"); 19 continue; 20 } 21 22 for(int i=1;i<=l;i++){ 23 ans[++p]=5+min(s,4); 24 s=s-min(s,4); 25 } 26 while(s){ 27 ans[++p]=min(s,4); 28 s=s-min(s,4); 29 } 30 for(int i=p;i>=1;i--){ 31 printf("%d",ans[i]); 32 } 33 printf("\n"); 34 } 35 return 0; 36 }
hdu 6023
水題，模擬。
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 int T,n,m,r,a,b,ans,cnt,pen[20]; 6 char c; 7 string s; 8 bool f[20]; 9 10 int main(){ 11 ios_base::sync_with_stdio(false),cin.tie(NULL),cout.tie(NULL); 12 cin>>T; 13 while(T--){ 14 cin>>n>>m; 15 memset(f,0,sizeof(f)); 16 memset(pen,0,sizeof(pen)); 17 ans=0; cnt=0; 18 while(m--){ 19 cin>>r>>a>>c>>b>>s; 20 r-=1000; 21 if(f[r]) continue; 22 if(s[0]=='A'){ 23 cnt++; 24 f[r]=1; 25 ans=ans+a*60+b+pen[r]; 26 } 27 else{ 28 pen[r]+=20; 29 } 30 } 31 cout<<cnt<<" "<<ans<<'\n'; 32 } 33 return 0; 34 }
hdu 6024
dp[i][j]為前i個的最小花費，j=1表示第i個建，j=0表示不建。轉移如下。其中dp[i][0]需要遍歷一遍前面的情況。
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 typedef long long ll; 5 struct node{ 6 ll x,c; 7 bool operator < (const node &t) const{ 8 return x<t.x; 9 } 10 }a[3003]; 11 12 ll n,dp[3003][3]; 13 14 int main(){ 15 while(~scanf("%lld",&n)){ 16 memset(dp,0,sizeof(dp)); 17 for(int i=1;i<=n;i++){ 18 scanf("%lld%lld",&a[i].x,&a[i].c); 19 } 20 sort(a+1,a+1+n); 21 dp[1][1]=a[1].c; 22 dp[1][0]=(1ll<<60); 23 for(int i=2;i<=n;i++){ 24 dp[i][1]=min(dp[i-1][0],dp[i-1][1])+a[i].c; 25 dp[i][0]=(1ll<<60); 26 ll tmp=0; 27 for(int j=i-1;j>=1;j--){ 28 tmp=tmp+(i-j)*(a[j+1].x-a[j].x); 29 dp[i][0]=min(dp[i][0],dp[j][1]+tmp); 30 } 31 } 32 printf("%lld\n",min(dp[n][0],dp[n][1])); 33 } 34 return 0; 35 }
hdu 6025
預處理前綴和后綴gcd，根據gcd的性質可以O(n)掃一遍得出答案。
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 const int maxn=1e5+20; 5 6 int n,T,a[maxn],pre[maxn],las[maxn],ans; 7 8 int main(){ 9 scanf("%d",&T); 10 while(T--){ 11 scanf("%d",&n); 12 for(int i=1;i<=n;i++){ 13 scanf("%d",&a[i]); 14 } 15 pre[1]=a[1]; las[n]=a[n]; 16 for(int i=2;i<=n;i++){ 17 pre[i]=__gcd(pre[i-1],a[i]); 18 } 19 for(int i=n-1;i>=1;i--){ 20 las[i]=__gcd(las[i+1],a[i]); 21 } 22 23 ans=max(pre[n-1],las[2]); 24 for(int i=2;i<=n-1;i++){ 25 ans=max(ans,__gcd(pre[i-1],las[i+1])); 26 } 27 printf("%d\n",ans); 28 } 29 return 0; 30 }
hdu 6026
稍微修改一下djkstra板子，記錄0到每個點的最短路有幾條，由於最后一定各保留一條最短路，相乘即為選擇的方案數。
1 #include<bits/stdc++.h> 2 #define inf 0x3f3f3f3f 3 #define miaojie1 4 #ifdef miaojie 5 #define dbg(args...) do {cout << #args << " : "; err(args);} while (0) 6 void err() {std::cout << std::endl;} 7 template<typename T, typename...Args> 8 void err(T a, Args...args){std::cout << a << ' '; err(args...);} 9 #else 10 #define dbg(...) 11 #endif 12 using namespace std; 13 typedef long long ll; 14 typedef pair<int,int>pii; 15 const ll mod=1e9+7; 16 const int maxn=3003; 17 18 struct edge{ 19 int nxt,to,cost; 20 }e[maxn]; 21 22 int n,head[55],d[55],cnt[55],p; 23 ll ans; 24 char s[55]; 25 bool vis[55]; 26 priority_queue<pii,vector<pii>,greater<pii> >q; 27 28 void addedge(int from,int to,int cost){ 29 e[p].nxt=head[from]; 30 e[p].to=to; 31 e[p].cost=cost; 32 head[from]=p++; 33 } 34 35 void DJ(){ 36 while(!q.empty()) q.pop(); 37 for(int i=0;i<n;i++){ 38 vis[i]=0; 39 d[i]=inf; 40 cnt[i]=1; 41 } 42 d[0]=0; 43 q.push(make_pair(0,0)); 44 while(!q.empty()){ 45 pii t=q.top(); 46 dbg(t.first,t.second); 47 q.pop(); 48 vis[t.second]=1; 49 for(int i=head[t.second];~i;i=e[i].nxt){ 50 if(vis[e[i].to]) continue; 51 if(d[e[i].to]>t.first+e[i].cost){ 52 d[e[i].to]=t.first+e[i].cost; 53 cnt[e[i].to]=1; 54 q.push(make_pair(d[e[i].to],e[i].to)); 55 } 56 else if(d[e[i].to]==t.first+e[i].cost) cnt[e[i].to]++; 57 } 58 } 59 } 60 61 int main(){ 62 while(~scanf("%d",&n)){ 63 memset(head,-1,sizeof(head)); p=0; ans=1; 64 for(int i=0;i<n;i++){ 65 scanf("%s",s); 66 for(int j=0;j<n;j++){ 67 if(s[j]!='0') addedge(i,j,s[j]-48); 68 } 69 } 70 DJ(); 71 for(int i=0;i<n;i++){ 72 ans=ans*cnt[i]%mod; 73 } 74 printf("%lld\n",ans); 75 } 76 return 0; 77 }
hdu 6027
水題，預處理一下暴力能過。
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 typedef long long ll; 5 const ll mod=1e9+7; 6 7 int T; 8 ll n,k,f[10002][7]; 9 10 void init(){ 11 for(int i=1;i<=10000;i++){ 12 f[i][0]=1; 13 for(int j=1;j<=5;j++){ 14 f[i][j]=f[i][j-1]*i%mod; 15 } 16 } 17 } 18 19 int main(){ 20 scanf("%d",&T); 21 init(); 22 while(T--){ 23 scanf("%lld%lld",&n,&k); 24 ll ans=0; 25 for(int i=1;i<=n;i++){ 26 ans=(ans+f[i][k])%mod; 27 } 28 printf("%lld\n",ans); 29 } 30 return 0; 31 }
hdu 6029
尋找是否存在完美匹配，從后往前貪心即可，盡量讓后面的連接。
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 int T,n,a[100005]; 6 7 int main(){ 8 scanf("%d",&T); 9 while(T--){ 10 scanf("%d",&n); 11 int cnt=0; 12 for(int i=2;i<=n;i++){ 13 scanf("%d",&a[i]); 14 if(a[i]==1) cnt++; 15 } 16 if(n%2||cnt<n/2){ 17 printf("No\n"); 18 continue; 19 } 20 21 int f=0; 22 bool jd=1; 23 for(int i=n;i>=2;i--){ 24 if(a[i]==2) f--; 25 else f++; 26 if(f<0){ 27 jd=0; 28 break; 29 } 30 } 31 if(jd) printf("Yes\n"); 32 else printf("No\n"); 33 } 34 return 0; 35 }
hdu 6030
紅記為1，藍記為0，題目條件轉化為連續2個或3個珠子中1的個數>=0的個數。考慮在2個連續的珠子后增加一個，有10-->101->01 , 01->011->11 , 11->111或110->11或10；
記末尾10的個數為f(a)，末尾01的個數為f(b)，末尾11的個數為f(c)，總個數f(n)；
則f(a+1)=f(c)，f(b+1)=f(a)，f(c+1)=f(b)+f(c)；
根據以上關系推出f(n)=f(n-1)+f(n-3)，顯然可以用矩陣快速冪解決。
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 typedef long long ll; 5 const int mod=1e9+7; 6 const int z = 3; 7 8 int T; 9 ll n; 10 struct ju{ 11 long long a[z][z]; 12 }; 13 14 ju muli(ju A,ju B){ 15 ju C; 16 for(int i=0;i<z;++i){ 17 for(int j=0;j<z;++j){ 18 C.a[i][j]=0; 19 for(int k=0;k<z;++k){ 20 C.a[i][j]+=A.a[i][k]*B.a[k][j]; 21 C.a[i][j]%=mod; 22 } 23 } 24 } 25 return C; 26 } 27 ju init(){ 28 ju res; 29 for(int i=0;i<z;++i){ 30 for(int j=0;j<z;++j){ 31 if(i==j)res.a[i][j]=1; 32 else res.a[i][j]=0; 33 } 34 } 35 return res; 36 } 37 ju pow(ju A,ll n){ 38 ju res=init(),temp=A; 39 for(;n;n/=2){ 40 if(n&1)res=muli(res,temp); 41 temp=muli(temp,temp); 42 } 43 return res; 44 } 45 void print(ju A){ 46 for(int i=0;i<z;++i){ 47 for(int j=0;j<z;++j){ 48 cout<<A.a[i][j]<<" "; 49 } 50 cout<<endl; 51 } 52 cout<<endl; 53 } 54 55 int main(){ 56 scanf("%d",&T); 57 while(T--){ 58 scanf("%lld",&n); 59 if(n==2) { 60 printf("3\n"); 61 continue; 62 } 63 if(n==3){ 64 printf("4\n"); 65 continue; 66 } 67 ju ans,t; 68 for(int i=0;i<z;i++){ 69 for(int j=0;j<z;j++){ 70 ans.a[i][j]=t.a[i][j]=0; 71 } 72 } 73 ans.a[0][0]=6; ans.a[1][0]=4; ans.a[2][0]=3; 74 t.a[0][0]=t.a[0][2]=t.a[1][0]=t.a[2][1]=1; 75 // print(t); 76 ans=muli(pow(t,n-4),ans); 77 // print(ans); 78 printf("%lld\n",ans.a[0][0]%mod); 79 } 80 return 0; 81 }
hdu 6032
在賈老板的指導下勉強混過了。
三個map分別存子串是否出現、出現次數、對應的sg。結構體存勝負，順便存分數，dfs跑sg的時候記得每次交換自己(p1)和對手(p2)。做着難受，I very vegetable.
1 #include<bits/stdc++.h> 2 #define inf 0x3f3f3f3f 3 #define miaojie1 4 #ifdef miaojie 5 #define dbg(args...) do {cout << #args << " : "; err(args);} while (0) 6 void err() {std::cout << std::endl;} 7 template<typename T, typename...Args> 8 void err(T a, Args...args){std::cout << a << ' '; err(args...);} 9 #else 10 #define dbg(...) 11 #endif 12 using namespace std; 13 14 struct node{ 15 int flag; 16 int p1,p2; 17 node(){} 18 node(bool _flag,int _p1,int _p2) : flag(_flag),p1(_p1),p2(_p2){} 19 bool operator < (const node &a)const{ 20 if(flag==a.flag){ 21 if(p2==a.p2){ 22 return p1<a.p1; 23 } 24 return p2>a.p2; 25 } 26 return flag<a.flag; 27 } 28 }; 29 30 map<string,int>jd; 31 map<string,set<int> >cnt; 32 map<string,node>sg; 33 int n; 34 char s[100]; 35 36 node dfs(string s){ 37 if(sg.count(s)){ 38 return sg[s]; 39 } 40 node t{1,inf,0}; 41 string ts; 42 for(char i='a';i<='z';i++){ 43 if(jd.count(s+i)) t=min(t,dfs(s+i)); 44 if(jd.count(i+s)) t=min(t,dfs(i+s)); 45 } 46 47 int sum=0,maxi=0; 48 for(int i=0;i<s.size();i++){ 49 sum=sum+s[i]-96; 50 maxi=max(maxi,s[i]-96); 51 } 52 sum=sum*maxi+cnt[s].size(); 53 dbg(s,t.flag,t.p1,t.p2,sum); 54 if(t.p1==inf){ 55 node tt(0,0,sum); 56 return tt; 57 } 58 sg[s].flag=1-t.flag; 59 sg[s].p1=t.p2; 60 sg[s].p2=t.p1+sum; 61 return sg[s]; 62 } 63 64 int main(){ 65 while(~scanf("%d",&n)){ 66 jd.clear(); cnt.clear(); sg.clear(); 67 for(int i=1;i<=n;i++){ 68 scanf("%s",&s); 69 int l=strlen(s); 70 for(int j=0;j<l;j++){ 71 string t; 72 for(int k=j;k<l;k++){ 73 t+=s[k]; 74 jd[t]=1; 75 cnt[t].insert(i); 76 } 77 } 78 } 79 node ans=dfs(""); 80 if(ans.flag) printf("Alice\n"); 81 else printf("Bob\n"); 82 printf("%d %d\n",ans.p1,ans.p2); 83 } 84 return 0; 85 }