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//通过观察，发现生成元在给出数n的n-100以内 //这么写不会超时 #include <iostream> #include <stdio.h> #include <stdlib.h> #include <cstring> #include <string.h> #include <cmath> #include <...

#include <stdio.h>                                 //隐藏错误：缺少头文件<string.h>#define maxn 10000000 + 10   //错误1：maxn太大，程序无法运行。int main(){    char s[maxn];    scanf("%s",s);    int tot = 0;    ...

Sample Input 4 1 2 3 5 1 5 5 2 1 3 3 3 1 2 3 0 0 Sample Output CASE# 1: 5 found at 4 CASE# 2: 2 not found 3 found at 3   题目大意是找到指定数字（经过排序后的）位置，用sort()和lower_bound()就可以做了，记住的是不能用...

Sample Input 10 move 9 onto 1 move 8 over 1 move 7 over 1 move 6 over 1 pile 8 over 6 pile 8 over 5 move 2 over 1 move 4 over 9 quit Sample Output 0: 0 1: 1 9 2 4 2: 3: 3 4: 5: 5 8 7 ...

Sample Input Adventures in Disneyland Two blondes were going to Disneyland when they came to a fork in the road. The sign read: "Disneyland Left." So they went home. Sample Output a adventures ...

Sample Input ladder came tape soon leader acme RIDE lone Dreis peat ScAlE orb eye Rides dealer NotE derail LaCeS drIed noel dire Disk mace Rob dries # Sample Output Disk NotE derail drIed ey...

Suppose you have to evaluate an expression like A*B*C*D*E where A,B,C,D and E are matrices. Since matrix multiplication is associative, the order in which multiplications are performed is arbitrary. ...

#include<bits/stdc++.h> using namespace std; int A[120]; bool isp[150]; int n; int vis[150]; void dfs(int cur){ if(cur==n && isp[A[0]+A[n-1]]){ for(int i = 0;i < n;i++) ...

用了4种方法实现： #include<bits/stdc++.h> using namespace std; typedef long long ll; const ll maxn=1e5+10; const ll mod=1e9+7; int a[10000],b[10000],c[10000]; int n = 1000,t1 = 0,t2 = 0,t3 = 0,t4 = 0...