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:

1. 1≤l1≤r1<l2≤r2<l3≤r3≤n

2. 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 (1≤T≤100), 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

2 
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=1;i<=9;i++){
        int ans1=s.find(goal.substr(0,i),0);
        if(ans1<0) continue;
        for(int j=1;j+i<=10;j++){
            int ans2=s.find(goal.substr(i,j),ans1+i);
            if(ans2<0) continue;
            int k=11-i-j;
            int ans3=s.find(goal.substr(i+j,k),ans2+j);
            if(ans3<0) continue;
            return true;
        }
    }
    return false;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        cin>>s;
        if(solve()){
            printf("YES
");
        }
        else{
            printf("NO
");
        }
    }
    return 0;
}