hdu 5311 Hidden String（find，substr）

Problem Description

Today is the 1st anniversary of BestCoder. Soda, the contest manager, gets a string s of length n. He wants to find three nonoverlapping substrings s[l1..r1], s[l2..r2], s[l3..r3] that:

. ≤l1≤r1<l2≤r2<l3≤r3≤n

. The concatenation of s[l1..r1], s[l2..r2], s[l3..r3] is "anniversary".

 

Input

There are multiple test cases. The first line of input contains an integer T (≤T≤), indicating the number of test cases. For each test case:

There's a line containing a string s (1≤|s|≤100) consisting of lowercase English letters.

 

Output

For each test case, output "YES" (without the quotes) if Soda can find such thress substrings, otherwise output "NO" (without the quotes).

 

Sample Input

annivddfdersewwefary
nniversarya

 

Sample Output

YES
NO

 

Source

BestCoder 1st Anniversary ($)

 

 题意：从s串中找出3段连续的字串组成“anniversary”

复习了下find，substr的用法，老是忘记

s.substr（i，j）表示从s串的i位置开始，长度为j的子串。

s.find（p，i），p为字串，表示从s串的i位置开始，寻找有没有等于p的子串，如果有返回s的首地址，否则返回-1

这题枚举“anniversary”的3个子串，在给出的s串中寻找就可以了！

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<queue>
 #include<cmath>
 #include<stdlib.h>
 #include<map>
 using namespace std;
 string s;
 string goal="anniversary";
 bool solve(){
     for(int i=;i<=;i++){
         int ans1=s.find(goal.substr(,i),);
         if(ans1<) continue;
         for(int j=;j+i<=;j++){
             int ans2=s.find(goal.substr(i,j),ans1+i);
             if(ans2<) continue;
             int k=-i-j;
             int ans3=s.find(goal.substr(i+j,k),ans2+j);
             if(ans3<) continue;
             return true;
         }
     }
     return false;
 }
 int main()
 {
     int t;
     scanf("%d",&t);
     while(t--){
         cin>>s;
         if(solve()){
             printf("YES\n");
         }
         else{
             printf("NO\n");
         }
     }
     return ;
 }