solved:9/13
upsolved:9/13
A.
數據范圍很小，bfs + 去重就能過。注意出口只有一端。
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 const int N = 6; 6 const int SEED = 31; 7 8 int a[N][N]; 9 int m; 10 set<uint64_t> f; 11 12 struct Car { 13 bool is_vertical; 14 int lx, ly, rx, ry; // or upos dpos 15 Car() : lx(INT_MAX), ly(INT_MAX), rx(0), ry(0) {} 16 } p[10 + 5]; 17 18 struct Stat { 19 int st[N][N]; 20 void copy() { 21 memcpy(st, a, sizeof(a)); 22 } 23 void paste() { 24 memcpy(a, st, sizeof(a)); 25 } 26 Stat() { memset(st, 0, sizeof(st)); } 27 }; 28 29 uint64_t encode() { 30 uint64_t seed = 0; 31 for (int i = 0; i < N; ++i) 32 for (int j = 0; j < N; ++j) 33 seed = (seed + a[i][j]) * SEED; 34 return seed; 35 } 36 37 void init_car() { 38 for (int i = 1; i <= m; ++i) 39 p[i] = Car(); 40 for (int i = 0; i < N; ++i) 41 for (int j = 0; j < N; ++j) { 42 int x = a[i][j]; 43 p[x].lx = min(p[x].lx, i); 44 p[x].ly = min(p[x].ly, j); 45 p[x].rx = max(p[x].rx, i); 46 p[x].ry = max(p[x].ry, j); 47 if (p[x].lx == p[x].rx) 48 p[x].is_vertical = false; 49 if (p[x].ly == p[x].ry) 50 p[x].is_vertical = true; 51 } 52 } 53 54 int bfs() { 55 queue<pair<int, Stat> > que; 56 Stat temp; 57 temp.copy(); 58 que.push(make_pair(0, temp)); 59 while (!que.empty()) { 60 int step = que.front().first; 61 if (step > 8) { 62 return -1; 63 } 64 if (a[2][5] == 1) { 65 return step + 2; 66 // cerr << "QWQ\n"; 67 } 68 temp = que.front().second; 69 que.pop(); 70 temp.paste(); 71 uint64_t code = encode(); 72 if (f.count(code)) 73 continue; 74 f.insert(code); 75 init_car(); 76 // do { 77 // cerr << "STEP: " << step << endl; 78 // for (int i = 0; i < N; ++i) { 79 // for (int j = 0; j < N; ++j) { 80 // cerr << a[i][j] << ' '; 81 // } 82 // cerr << endl; 83 // } 84 // cerr << endl; 85 // } while (false); 86 for (int t = 1; t <= m; ++t) { 87 if (p[t].is_vertical) { 88 int lx = p[t].lx; 89 int rx = p[t].rx; 90 int y = p[t].ly; 91 if (0 < lx && a[lx - 1][y] == 0) { 92 a[lx - 1][y] = t; 93 a[rx][y] = 0; 94 temp.copy(); 95 que.push(make_pair(step + 1, temp)); 96 a[rx][y] = t; 97 a[lx - 1][y] = 0; 98 } 99 if (rx + 1 < N && a[rx + 1][y] == 0) { 100 a[rx + 1][y] = t; 101 a[lx][y] = 0; 102 temp.copy(); 103 que.push(make_pair(step + 1, temp)); 104 a[lx][y] = t; 105 a[rx + 1][y] = 0; 106 } 107 } else { 108 int x = p[t].lx; 109 int ly = p[t].ly; 110 int ry = p[t].ry; 111 if (0 < ly && a[x][ly - 1] == 0) { 112 a[x][ly - 1] = t; 113 a[x][ry] = 0; 114 temp.copy(); 115 que.push(make_pair(step + 1, temp)); 116 a[x][ly - 1] = 0; 117 a[x][ry] = t; 118 } 119 if (ry + 1 < N && a[x][ry + 1] == 0) { 120 a[x][ry + 1] = t; 121 a[x][ly] = 0; 122 temp.copy(); 123 que.push(make_pair(step + 1, temp)); 124 a[x][ry + 1] = 0; 125 a[x][ly] = t; 126 } 127 } 128 } 129 } 130 return -1; 131 } 132 133 int main() { 134 ios::sync_with_stdio(false); 135 for (int i = 0; i < N; ++i) 136 for (int j = 0; j < N; ++j) { 137 cin >> a[i][j]; 138 m = max(a[i][j], m); 139 } 140 init_car(); 141 cout << bfs() << endl; 142 }
C.
簽到題
1 #include<bits/stdc++.h> 2 using namespace std; 3 int n; 4 int a[55]; 5 int main() 6 { 7 scanf("%d",&n); 8 for(int i=1;i<=n;++i)scanf("%d",&a[i]); 9 bool yes=1; 10 for(int i=1;i<=n;++i) 11 for(int j=1;j<=n;++j)if(j!=i) 12 for(int k=1;k<=n;++k)if(k!=i&&k!=j) 13 { 14 int x=abs(a[i]-a[j]); 15 if(x%a[k])yes=0; 16 } 17 if(yes)puts("yes"); 18 else puts("no"); 19 }
D.
簽到題
1 #include<bits/stdc++.h> 2 using namespace std; 3 string s; 4 int main() 5 { 6 bool has=0; 7 while(cin>>s) 8 { 9 if(s=="bubble"||s=="tapioka")continue; 10 has=1; 11 cout<<s<<" "; 12 } 13 if(!has)puts("nothing"); 14 }
E.
構造一個長度1999的序列
考慮第一個數是負數，那么假算法一定不會包含這個數,取\(a_1=-1\)
序列為\(-1,a_2,a_3,…,a_{1999}\)
那么假算法的解為\(1998*\sum_{i=2}^{1999}{a_i}\)
真算法的解為\(1999*(\sum_{i=2}^{1999}{a_i}-1)\)
那么假算法的解比真算法小$k$，則有\(1999*(\sum_{i=2}^{1999}{a_i}-1)=1998*\sum_{i=2}^{1999}{a_i}+k\)
則有\(\sum_{i=2}^{1999}{a_i}=1999+k\)
隨便構造一下湊夠\(k\)就完事了
1 #include<bits/stdc++.h> 2 using namespace std; 3 int T,k,L; 4 int main() 5 { 6 scanf("%d",&T); 7 while(T--) 8 { 9 scanf("%d%d",&k,&L); 10 if(L>=2000)puts("-1"); 11 else 12 { 13 puts("1999"); 14 printf("-1 "); 15 int s=k+1999,x=1000000; 16 for(int i=1;i<=1998;++i) 17 { 18 if(s>=x) 19 { 20 printf("%d ",x); 21 s-=x; 22 } 23 else 24 { 25 printf("%d ",s); 26 s=0; 27 } 28 } 29 } 30 } 31 }
H.
考慮方程\(\frac{1}{x}+\frac{1}{y}=\frac{1}{n}\)
移項通分有\(\frac{1}{y}=\frac{x-n}{xn}\)
則\(y=\frac{xn}{x-n}\)
令\(a=x-n\)，則有\(y=n+\frac{n^2}{a}\)
那么枚舉\(n^2\)的約數,解出\(x,y\)即可
然后\(a=x \oplus y\)
1 #include<bits/stdc++.h> 2 #define ll long long 3 using namespace std; 4 int T; 5 ll n; 6 int main() 7 { 8 cin>>T; 9 while(T--) 10 { 11 cin>>n; 12 ll ans=0; 13 for(ll i=1;i<=n;++i)if((n*n)%i==0) 14 { 15 ll y=n+n*n/i; 16 ll x=n+i; 17 ll a=x^y; 18 ans=max(ans,a); 19 } 20 cout<<ans<<endl; 21 } 22 }
J.
二進制枚舉集合，簽到題
1 #include<bits/stdc++.h> 2 using namespace std; 3 int T,n,m; 4 bitset<505> a[18]; 5 int main() 6 { 7 scanf("%d",&T); 8 while(T--) 9 { 10 scanf("%d%d",&n,&m); 11 for(int i=0;i<m;++i) 12 { 13 char str[505]; 14 scanf("%s",str); 15 for(int j=0;j<n;++j)a[i][j]=0; 16 for(int j=0;j<n;++j)if(str[j]=='1')a[i][j]=1; 17 } 18 int ans=m+1; 19 for(int S=0;S<(1<<m);++S) 20 { 21 bitset<505> b; 22 for(int i=0;i<n;++i)b[i]=0; 23 int cnt=0; 24 for(int i=0;i<m;++i)if(S&(1<<i))b|=a[i],cnt++; 25 bool yes=1; 26 for(int i=0;i<n;++i)if(!b[i])yes=0; 27 if(yes)ans=min(ans,cnt); 28 } 29 if(ans<=m)printf("%d\n",ans); 30 else puts("-1"); 31 } 32 }
K.
簽到題。合並石子，優先隊列即可。
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 int main() { 6 ios::sync_with_stdio(false); 7 int T; 8 cin >> T; 9 while (T--) { 10 int n; 11 cin >> n; 12 priority_queue<int, vector<int>, greater<int>> que; 13 for (int i = 0; i < n; ++i) { 14 int x; 15 cin >> x; 16 que.push(x); 17 } 18 int sum = 0; 19 while (!que.empty()) { 20 if (que.size() == 1) { 21 cout << sum << endl; 22 break; 23 } 24 int x = que.top(); 25 que.pop(); 26 int y = que.top(); 27 que.pop(); 28 sum += x + y; 29 que.push(x + y); 30 } 31 } 32 }
L.
因為可能是退化的四邊形,所以不能在凸包上直接找點
考慮四邊形的對角線一定是凸包上的對踵點，那么這步可以旋轉卡殼得到對踵點
我們下面不能直接在凸包上三分了，但本題允許\(O(n^2)\)的時間復雜度，那么我們枚舉點
四邊形可能是凹的，所以我們枚舉的時候需要分別記錄這條線兩邊最大最小的面積
如果某邊沒有點，就是最大減最小，否則就是兩邊相加
1 #include<bits/stdc++.h> 2 #define ll long long 3 using namespace std; 4 struct Point 5 { 6 ll x,y; 7 Point(ll X=0,ll Y=0):x(X),y(Y){} 8 }; 9 typedef Point Vector; 10 inline Vector operator + (Vector A,Vector B){return Vector(A.x+B.x,A.y+B.y);} 11 inline Vector operator - (Point A,Point B){return Vector(A.x-B.x, A.y-B.y);} 12 ll Cross(Vector A,Vector B){return A.x*B.y-A.y*B.x;} 13 bool operator < (Point A,Point B){if(A.x==B.x)return A.y<B.y;return A.x<B.x;} 14 bool operator == (Point A,Point B){return (A.x==B.x)&&(A.y==B.y);} 15 vector<Point> ConvexHull(vector<Point>& p) 16 { 17 sort(p.begin(),p.end()); 18 p.erase(unique(p.begin(),p.end()),p.end()); 19 int n=p.size(); 20 int m=0; 21 vector<Point> ch(n+1); 22 for(int i=0;i<n;++i) 23 { 24 while(m>1&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)m--; 25 ch[m++]=p[i]; 26 } 27 int k=m; 28 for(int i=n-2;i>=0;--i) 29 { 30 while(m>k&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)m--; 31 ch[m++]=p[i]; 32 } 33 if(n>1)m--; 34 ch.resize(m); 35 return ch; 36 } 37 int T,n,m; 38 vector<Point> A,H; 39 ll calc(Point U,Point V,bool tst=0) 40 { 41 bool firU=0,firV=0; 42 vector<Point> B;B.clear(); 43 for(Point p:A) 44 { 45 if(p==U&&(!firU))firU=1; 46 else if(p==V&&(!firV))firV=1; 47 else B.push_back(p); 48 } 49 ll mx1=-1,mx2=-1,mn1=(ll)2e18,mn2=(ll)2e18; 50 for(Point p:B) 51 { 52 //if(tst)printf("(%I64d,%I64d)\n",p.x,p.y); 53 if(Cross(V-U,p-U)<0) 54 { 55 mx1=max(mx1,abs(Cross(V-U,p-U))); 56 mn1=min(mn1,abs(Cross(V-U,p-U))); 57 } 58 else 59 { 60 mx2=max(mx2,abs(Cross(V-U,p-U))); 61 mn2=min(mn2,abs(Cross(V-U,p-U))); 62 } 63 } 64 if(mx1==-1)return mx2-mn2; 65 else if(mx2==-1)return mx1-mn1; 66 else return mx1+mx2; 67 } 68 int main() 69 { 70 scanf("%d",&T); 71 while(T--) 72 { 73 A.clear(); 74 scanf("%d",&n); 75 for(int i=1;i<=n;++i) 76 { 77 ll x,y; 78 scanf("%I64d%I64d",&x,&y); 79 A.push_back(Point(x,y)); 80 } 81 H=A; 82 H=ConvexHull(H); 83 m=H.size(); 84 if(m<=2) 85 { 86 puts("0"); 87 continue; 88 } 89 int j=1; 90 ll ans=0; 91 for(int i=0;i<m;++i) 92 { 93 while(abs(Cross(H[j]-H[i],H[j]-H[(i+1)%m]))<abs(Cross(H[(j+1)%m]-H[i],H[(j+1)%m]-H[(i+1)%m])))j=(j+1)%m; 94 ans=max(ans,calc(H[i],H[j])); 95 } 96 if(ans&1)printf("%I64d.5\n",ans/2); 97 else printf("%I64d\n",ans/2); 98 } 99 }
M.
題意為求解$\dfrac{\binom{N}{M}}{D^k}$，其中$\binom{N}{M}$恰好整除$D^k$。
對$D$質因數分解，得$D=\prod_{i=0}p_i^{a_i}$。我們可以單獨求得每個$p^a$的答案，再用 CRT 合並即可。
當對於$p^a$求答案時，有：
$$ \binom{N}{M}=\frac{N!}{M!(N-M)!}=\frac{(N!/p^x)}{(M!)/p^y((N-M)!)/p^z}\cdot p^{-x+y+z} $$
答案為$\frac{(N!/p^x)}{(M!)/p^y((N-M)!)/p^z}$。考慮如何計算這個式子。
令$f(n)=\frac{n!}{p^h}$，其中$p^h\mid n!$且$p^{h+1}\nmid n!$，$g(n)=\prod_{i=1,d\nmid i}^{n}i$。類似 exlucas 方法，有
$$f(n)=f(\frac{n}{p})\cdot g(p)^{\frac{n}{p}}\cdot g(n\bmod p)$$
所以答案為
$$\dfrac{\binom{N}{M}}{D^k}=\dfrac{\binom{N}{M}/p^h\cdot p^{h-ka}}{\Big(\frac{D}{p^a}\Big)^k}$$
合並即可。
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 const int MAXD = 16000000 + 10; 6 const int MAXN = 100000 + 10; 7 8 int64_t qpow(int64_t a, int64_t b, int64_t p) { 9 int64_t ans = 1; 10 while (b) { 11 if (b & 1) 12 ans = ans * a % p; 13 a = a * a % p; 14 b >>= 1; 15 } 16 return ans; 17 } 18 19 void exgcd(int64_t a, int64_t b, int64_t& x, int64_t& y, int64_t& d) { 20 if (!b) 21 d = a, x = 1, y = 0; 22 else 23 exgcd(b, a % b, y, x, d), y -= a / b * x; 24 } 25 26 int64_t inv(int64_t a, int64_t p) { 27 int64_t x, y, d; 28 exgcd(a, p, x, y, d); 29 x = (x % p + p) % p; 30 return x; 31 } 32 33 34 int64_t N, M, D, K; 35 int64_t a[MAXN], mi[MAXN], pi[MAXN], num[MAXN], ki[MAXN]; 36 int64_t n, g[MAXD]; 37 38 int64_t _cal(int64_t x, int64_t p) { 39 int64_t ans = 0; 40 while (x) { 41 ans += x / p; 42 x /= p; 43 } 44 return ans; 45 } 46 47 int64_t calc(int64_t p) { 48 return _cal(N, p) - _cal(M, p) - _cal(N - M, p); 49 } 50 int64_t CRT() { 51 int64_t Mi = 1, ans = 0, x, y, d; 52 for (int i = 1; i <= n; ++i) 53 Mi *= mi[i]; 54 for (int i = 1; i <= n; ++i) { 55 exgcd(Mi / mi[i], mi[i], x, y, d); 56 x = (x % mi[i] + mi[i]) % mi[i]; 57 ans = (ans + Mi / mi[i] * x % Mi * a[i] % Mi) % Mi; 58 } 59 return ans; 60 } 61 62 void factorize(int64_t x) { 63 for (int64_t i = 2; i * i <= x; ++i) { 64 if (x % i == 0) { 65 num[++n] = 0; 66 pi[n] = i; 67 mi[n] = 1; 68 while (x % i == 0) { 69 ++num[n]; 70 x /= i; 71 mi[n] *= i; 72 } 73 } 74 } 75 if (x > 1) { 76 num[++n] = 1; 77 pi[n] = x; 78 mi[n] = x; 79 } 80 } 81 82 int64_t fact(int64_t n, int64_t x, int64_t p) { 83 if (!n) 84 return 1; 85 int64_t ans = qpow(g[p], n / p, p) * g[n % p] % p; 86 return ans * fact(n / x, x, p) % p; 87 } 88 89 int64_t C(int64_t x, int64_t p) { 90 g[0] = 1; 91 for (int64_t i = 1; i <= p; ++i) { 92 g[i] = g[i - 1]; 93 if (i % x) 94 g[i] = g[i] * i % p; 95 } 96 return fact(N, x, p) * inv(fact(M, x, p), p) % p 97 * inv(fact(N - M, x, p), p) % p; 98 } 99 100 int main() { 101 ios::sync_with_stdio(false); 102 cin >> N >> M >> D; 103 factorize(D); 104 K = LLONG_MAX; 105 for (int i = 1; i <= n; ++i) { 106 ki[i] = calc(pi[i]); 107 K = min(K, ki[i] / num[i]); 108 } 109 for (int i= 1; i <= n; ++i) { 110 a[i] = C(pi[i], mi[i]) * qpow(pi[i], ki[i] - K * num[i], mi[i]) % mi[i] 111 * qpow(inv(D / mi[i], mi[i]), K, mi[i]) % mi[i]; 112 } 113 cout << CRT() << endl; 114 }