1.题目描述：点击打开链接
2.解题思路：本题要求在一个串中找出5个等距离的‘*’，如果有，输出yes，否则输出no。思路很清晰，，首先枚举步长len，然后枚举起点i，判断是否能够连续不间断地跳跃5次且均为‘*’即可。本题还可以加速寻找，先寻找是否存在‘*****’，如果有，直接输出yes。
3.代码：
#define _CRT_SECURE_NO_WARNINGS #include<iostream>#include<algorithm>#include<string>#include<sstream>#include<set>#include<vector>#include<stack>#include<map>#include<queue>#include<deque>#include<cstdlib>#include<cstdio>#include<cstring>#include<cmath>#include<ctime>#include<functional>using namespace std;string s;int main(){//freopen("t.txt", "r", stdin);int n;while (~scanf("%d", &n)){cin >> s;int i;for (i = 0; i < n; i++)if (s[i] == '*')//找到第一个‘*’的位置break;int cnt, ok = 0;if (strstr(s.c_str(), "*****") != NULL)ok = 1;//加速措施else{for (int len = 1; len < n; len++)//枚举步长{int cnt, k;for (int j = i; j < n – 4 * len; j++)//枚举起点{cnt = 0, k = 0;while (k < 5){if (s[j + k*len] == '*')cnt++;else break;k++;}if (cnt == 5){ ok = 1; break; }}}}if (ok)puts("yes");else puts("no");}return 0;}
就是对虚怀若谷谦虚谨慎八个字真正理解的人，
