hdoj 5311 Hidden String (暴力枚舉)

Hidden String

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1950    Accepted Submission(s): 724

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

 

題意：在給出的字符串裏能否找到三段不重疊的子串，使它們組成單詞"anniversary"，能輸出YES 不能輸出NO

思路：數據到100 ，暴力枚舉，使用string類方便些，每次枚舉單詞"anniversary"的三個子串，然後在給出的串裏看能否找到

code

#include <iostream>
#include<cstdio>
#include<string>
using namespace std;

int main()
{
    string s,s1,s2,s3,str="anniversary";
    int t;
    scanf("%d",&t);
    while(t--)
    {
        cin>>s;
        int len_s=s.length();
        int len_str=str.length();
        int flag=0;
        for(int i=1;i<=len_str-2;i++)
        {
            for(int j=1;j<=len_str-i-1;j++)
            {
                s1=str.substr(0,i);
                s2=str.substr(i,j);
                s3=str.substr(i+j,len_str-i-j);
                int st1=s.find(s1,0);
                if(st1==-1)
                    continue;
                int st2=s.find(s2,st1+i);
                if(st2==-1)
                    continue;
                int st3=s.find(s3,st2+j);
                if(st3==-1)
                    continue;
                flag=1;
                break;
            }
            if(flag)
                break;
        }
        if(flag)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}