A,C,I簽到題，只搞了8題，還一題是神仙做的，我不會
鏈接：https://www.nowcoder.com/acm/contest/122/B
來源：牛客網
取石子
時間限制：C/C++ 1秒，其他語言2秒
空間限制：C/C++ 32768K，其他語言65536K
64bit IO Format: %lld
空間限制：C/C++ 32768K，其他語言65536K
64bit IO Format: %lld
題目描述
現在有兩堆石子，兩個人輪流從中取石子，且每個人每一次只能取1、3或9個石子，取到最后一個石子的人win。
假設先手后手都會選擇最好的方式來取石子，請您判斷先后手的輸贏情況。
假設先手后手都會選擇最好的方式來取石子，請您判斷先后手的輸贏情況。
輸入描述:
多組輸入
每組一行，一行包括兩個正整數n1和n2(1<=n1<=100,1<=n2<=100)，代表了兩堆石子的數目
輸出描述:
如果先手能贏，輸出"win"；否則就輸出"lose"。
示例1
輸入
1 1 1 2
輸出
lose win
分析：這題挺有搞笑，兩堆加起來是奇數就win,否則lose。
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鏈接：https://www.nowcoder.com/acm/contest/122/D
來源：牛客網
小C的記事本
時間限制：C/C++ 2秒，其他語言4秒
空間限制：C/C++ 131072K，其他語言262144K
64bit IO Format: %lld
空間限制：C/C++ 131072K，其他語言262144K
64bit IO Format: %lld
題目描述
小C最近學會了java小程序的開發，他很開心，於是想做一個簡單的記事本程序練練手。
他希望他的記事本包含以下功能：
1、append(str)，向記事本插入字符串 str（英文字符）
2、delete(k)，刪除記事本最后k個字符（保證不為空串）
3、print(k)，輸出記事本第k個字符（保證不為空串）
4、undo()，撤銷最近的1（或者）操作，使記事本回到1（或者2）操作之前的狀態
可憐的小C琢磨了半天還是做不來，聰明的你能解決小C的問題嗎？
他希望他的記事本包含以下功能：
1、append(str)，向記事本插入字符串 str（英文字符）
2、delete(k)，刪除記事本最后k個字符（保證不為空串）
3、print(k)，輸出記事本第k個字符（保證不為空串）
4、undo()，撤銷最近的1（或者）操作，使記事本回到1（或者2）操作之前的狀態
可憐的小C琢磨了半天還是做不來，聰明的你能解決小C的問題嗎？
輸入描述:
多組輸入
第一行輸入一個整數q，代表操作總數
以下q行每行描述了一個操作，每行以一個整數t開始（1 <= t <= 4）。
t表示上述問題陳述中定義的操作類型。 如果操作需要參數，則后跟空格分隔的參數。
題目保證所有操作均合法
1 <= q <= 10^6
1 <= k <= |記事本內容長度|
每個測試數據中str的總長度 <= 10^6
請使用 ios::sync_with_stdio(false); 對讀寫進行加速
輸出描述:
所有操作類型3必須輸出第k個字符，每行以換行符結束。
示例1
輸入
8 1 ab 3 2 2 2 1 cd 3 1 4 4 3 1
輸出
b c a
說明
**樣例解釋**
假設記事本用字符串S表示
1、插入ab，S="ab"
2、輸出第2個字符，是b
3、刪除最后2個字符，S=""
4、插入cd, S="cd"
5、輸出第1個字符，是c
6、撤銷，此時S=""
7、撤銷，此時S="ab"
8、輸出第1個字符，是a
分析：這題很實用啊，現在都很喜歡回到上一步，和武大校賽那題有點像，不過這里是用stack解決的，因為可以回到之前所有的步驟，
和電腦上ctrl+z有點像,每次保存/刪除輸入字符串，str += str1，對於1,2進行一步操作就壓棧,最后s.pop(),即可
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JgGgBQDOSJss77r 67 :r;::: , PH r, ,;5: ,::::;7Ls7Lc7;: ,:7JP17rJUs 68 .:J: :; .,:K6BQS7:.,.,. 69 :r7Ji;r7;::;;. 70 . **/ 71 72 #include <bits/stdc++.h> 73 using namespace std; 74 75 #define FFI(a, b) for(int i = a; i < b; i++) 76 #define FFJ(a, b) for(int j = a; j < b; j++) 77 #define RR(a, b) for(int i = a; i > b; i++) 78 #define ME(a, b) memset(a, b, sizeof(a)) 79 #define SC(x) scanf("%d", &x) 80 #define PR(x) printf("%d\n", x) 81 #define INF 0x3f3f3f3f 82 #define MAX 1001 83 #define MOD 1000000007 84 #define E 2.71828182845 85 #define M 8 86 #define N 6 87 typedef long long LL; 88 const double PI = acos(-1.0); 89 typedef pair<int, int> Author; 90 vector<pair<string, int> > VP; 91 92 int main(void){ 93 #ifdef LOCAL 94 freopen("in.txt", "r", stdin); 95 freopen("out.txt", "w", stdout); 96 #endif 97 ios::sync_with_stdio(false); cin.tie(0); 98 int n, command, temp;string str, str1; 99 100 while(cin>>n){ 101 stack<string> s; str = "";s.push(str); 102 while(n--){ 103 cin>>command; 104 if(command == 1){ 105 cin>>str1; 106 str += str1; 107 s.push(str); 108 } 109 else if(command == 2){ 110 cin>>temp; 111 str.erase(str.size() - temp, temp); 112 s.push(str); 113 } 114 else if(command == 3){ 115 cin>>temp; 116 cout<<str[temp - 1]<<endl; 117 } 118 else{ 119 s.pop(); 120 str = s.top(); 121 } 122 } 123 } 124 125 return EXIT_SUCCESS; 126 }
鏈接：https://www.nowcoder.com/acm/contest/122/F
來源：牛客網
阿湯的疑惑
時間限制：C/C++ 1秒，其他語言2秒
空間限制：C/C++ 32768K，其他語言65536K
64bit IO Format: %lld
空間限制：C/C++ 32768K，其他語言65536K
64bit IO Format: %lld
題目描述
阿湯同學最近剛學數論，他發現數論實在是太有趣了，於是他想讓你也感受一下數論的樂趣。現在他給你一個正整數 N 和一個正整數 M，要求你用 N 對 M 進行取余操作，即 N % M，記余數為 S。
但是他發現這樣好像並不能讓你感受到數論的樂趣，於是他想讓你在N 對 M 取余操作的基礎上再求出這個余數 S 能分解出多少個不同質因數。
質因數：質因數在數論里是指能整除給定正整數的質數，質數就是只能整除 1 和本身的數，定義 2 是最小的質數。
但是他發現這樣好像並不能讓你感受到數論的樂趣，於是他想讓你在N 對 M 取余操作的基礎上再求出這個余數 S 能分解出多少個不同質因數。
質因數：質因數在數論里是指能整除給定正整數的質數，質數就是只能整除 1 和本身的數，定義 2 是最小的質數。
輸入描述:
從標准輸入讀入數據。
輸入包含多組數據，第一行一個整數 T 代表數據組數。接下來依
次描述每組數據，對於每組數據：
第一行輸入正整數 N，第二行輸入正整數 M
【數據規模】
1≤N≤10^100
1≤M≤2^31-1
輸出描述:
輸出到標准輸出。
對於每組數據，輸出一行：
余數 S 能分解出的不同質因數的個數。
示例1
輸入
2 68 40 6 180
輸出
2 2
分析：現在題目都好考思維了，求余數簡單，題目就是求s有多少素數，它可以整除.
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JgGgBQDOSJss77r 67 :r;::: , PH r, ,;5: ,::::;7Ls7Lc7;: ,:7JP17rJUs 68 .:J: :; .,:K6BQS7:.,.,. 69 :r7Ji;r7;::;;. 70 . **/ 71 72 #include <bits/stdc++.h> 73 using namespace std; 74 75 #define FFI(a, b) for(int i = a; i < b; i++) 76 #define FFJ(a, b) for(int j = a; j < b; j++) 77 #define RR(a, b) for(int i = a; i > b; i++) 78 #define ME(a, b) memset(a, b, sizeof(a)) 79 #define SC(x) scanf("%d", &x) 80 #define PR(x) printf("%d\n", x) 81 #define INF 0x3f3f3f3f 82 #define MAX 1001 83 #define MOD 1000000007 84 #define E 2.71828182845 85 #define M 8 86 #define N 6 87 typedef long long LL; 88 const double PI = acos(-1.0); 89 typedef pair<int, int> Author; 90 vector<pair<string, int> > VP; 91 92 int main(void){ 93 #ifdef LOCAL 94 freopen("in.txt", "r", stdin); 95 freopen("out.txt", "w", stdout); 96 #endif 97 ios::sync_with_stdio(false); cin.tie(0); 98 LL T, len, n, m, n1, res, ans;string str; 99 100 cin>>T; 101 while(T--){ 102 cin>>str>>m;n = res = 0; 103 FFI(0, str.size())n = (n * 10 + str[i] - '0') % m; 104 n1 = (LL)sqrt(n); 105 106 //ÖÊÒòÊý 107 FFI(2, n1 + 1)if(n % i == 0){res++;while(n % i == 0) n /= i;} 108 109 if(n != 1)res++; 110 cout<<res<<endl; 111 } 112 113 return EXIT_SUCCESS; 114 }
鏈接：https://www.nowcoder.com/acm/contest/122/G
來源：牛客網
阿湯的數組
時間限制：C/C++ 1秒，其他語言2秒
空間限制：C/C++ 32768K，其他語言65536K
64bit IO Format: %lld
空間限制：C/C++ 32768K，其他語言65536K
64bit IO Format: %lld
題目描述
阿湯同學為了准備下學期的 ACM-ICPC，刷了很多的題目，他覺得自己已經比較厲害了，於是想出個題目考考你。現在他給你一個數組 A，問你是否能將該數組划分成數組 B、C 使得 B 數組的平均數和C 數組的平均數相等，數組 B 和 C 都不能為空。
輸入描述:
從標准輸入讀入數據。
輸入包含多組數據，第一行一個整數 T 代表數據組數。接下來依次描述每組數據，對於每組數據：
第一行輸入正整數 N，第二行輸入 N 個非負整數
1≤|A|≤30 （數組 A 的長度范圍在 1 到 30 之間 ）
0≤A[i]≤10000 （數組 A 中的元素）
輸出描述:
輸出到標准輸出。
對於每組數據，輸出一行：
如果能划分成滿足題目要求的數組 B 和 C 則輸出 yes，否則輸出no
示例1
輸入
1 8 1 2 3 4 5 6 7 8
輸出
yes
說明
樣例說明：將數組 A 划分成【1,4,5,8】和【2,3,6,7】,平均數為 4.5
分析：DP，板子題
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 int t,n; 5 bool dp[310005][33]; 6 int s[35]; 7 int solve(){ 8 cin>>n; 9 double sum=0; 10 for(int i=1;i<=n;i++){ 11 cin>>s[i]; 12 sum+=s[i]; 13 } 14 sum/=n; 15 memset(dp,0,sizeof(dp)); 16 int maxn=0; 17 dp[0][0]=1; 18 for(int i=1;i<=n;i++){ 19 for(int j=maxn;j>=0;j--){ 20 for(int z=0;z<n-1;z++){ 21 if(dp[j][z]){ 22 int k=j+s[i]; 23 dp[k][z+1]=1; 24 maxn=max(maxn,k); 25 if(fabs(k*1.0/(z+1)-sum)<=0.0000000001){ 26 // cout<<k<<" "<<z+1<<endl; 27 // cout<<sum<<endl; 28 return 1; 29 } 30 } 31 } 32 } 33 } 34 return 0; 35 } 36 int main() 37 { 38 int t; 39 cin>>t; 40 while(t--){ 41 if(solve()) cout<<"yes"<<endl; 42 else cout<<"no"<<endl; 43 } 44 return 0; 45 }
鏈接：https://www.nowcoder.com/acm/contest/122/H
來源：牛客網
小q的數列
時間限制：C/C++ 1秒，其他語言2秒
空間限制：C/C++ 32768K，其他語言65536K
64bit IO Format: %lld
空間限制：C/C++ 32768K，其他語言65536K
64bit IO Format: %lld
題目描述
小q最近迷上了各種好玩的數列，這天，他發現了一個有趣的數列，其遞推公式如下：
f[0]=0 f[1]=1;
f[i]=f[i/2]+f[i%2];(i>=2)
現在，他想考考你，問：給你一個n，代表數列的第n項，你能不能馬上說出f[n]的值是多少，以及f[n]所代表的值第一次出現在數列的哪一項中？（這里的意思是：可以發現這個數列里某幾項的值是可能相等的，則存在這樣一個關系f[n'] = f[n] = f[x/2]+f[x%2] = f[x]...（n'<n<x） 他們的值都相等，這里需要你輸出最小的那個n'的值)(n<10^18)
f[0]=0 f[1]=1;
f[i]=f[i/2]+f[i%2];(i>=2)
現在，他想考考你，問：給你一個n，代表數列的第n項，你能不能馬上說出f[n]的值是多少，以及f[n]所代表的值第一次出現在數列的哪一項中？（這里的意思是：可以發現這個數列里某幾項的值是可能相等的，則存在這樣一個關系f[n'] = f[n] = f[x/2]+f[x%2] = f[x]...（n'<n<x） 他們的值都相等，這里需要你輸出最小的那個n'的值)(n<10^18)
輸入描述:
輸入第一行一個t
隨后t行，每行一個數n，代表你需要求數列的第n項，和相應的n'
(t<4*10^5)
輸出描述:
輸出每行兩個正整數
f[n]和n'，以空格分隔
示例1
輸入
2 0 1
輸出
0 0 1 1
分析：這題我本來也是想找循環環，但是發現有點問題，后來再用n / 2 ,n % 2想的
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JgGgBQDOSJss77r 67 :r;::: , PH r, ,;5: ,::::;7Ls7Lc7;: ,:7JP17rJUs 68 .:J: :; .,:K6BQS7:.,.,. 69 :r7Ji;r7;::;;. 70 . **/ 71 72 #include <bits/stdc++.h> 73 using namespace std; 74 75 #define FFI(a, b) for(int i = a; i < b; i++) 76 #define FFJ(a, b) for(int j = a; j < b; j++) 77 #define RR(a, b) for(int i = a; i > b; i++) 78 #define ME(a, b) memset(a, b, sizeof(a)) 79 #define SC(x) scanf("%d", &x) 80 #define PR(x) printf("%d\n", x) 81 #define INF 0x3f3f3f3f 82 #define MAX 1001 83 #define MOD 1000000007 84 #define E 2.71828182845 85 #define M 8 86 #define N 6 87 typedef long long LL; 88 const double PI = acos(-1.0); 89 typedef pair<int, int> Author; 90 vector<pair<string, int> > VP; 91 92 int main(void){ 93 // #ifdef LOCAL 94 // freopen("in.txt", "r", stdin); 95 // freopen("out.txt", "w", stdout); 96 // #endif 97 // ios::sync_with_stdio(false); cin.tie(0); 98 LL T, n, ans, len; 99 100 cin>>T; 101 while(T--){ 102 scanf("%lld", &n); len = ans = 0; 103 while(n > 1){ 104 if(n % 2 == 1) len++; 105 n /= 2; 106 } 107 len += n == 0 ? 0 : 1; 108 FFI(0, len) ans = ans * 2 + 1; 109 printf("%lld %lld\n", len, ans); 110 } 111 112 return EXIT_SUCCESS; 113 }