HDU 5311 Hidden String (优美的暴力)

Hidden String

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

Total Submission(s): 52    Accepted Submission(s): 25

Problem Description

Today is the 1st anniversary of BestCoder. Soda, the contest manager, gets a strings 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 integerT(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 ($)

 
题目链接：http://acm.hdu.edu.cn/showproblem.php?

pid=5311

题目大意：问在一个字符串里能不能找到连续三个区间拼成anniversary

题目分析：暴力，枚举每段的长度

#include <cstdio>
#include <cstring>
char s[200], con[] = "anniversary";

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%s", s);
        int len = strlen(s);
        bool flag = false;
        for(int i = 0; i <= 8; i++)
        {
            for(int j = i + 1; j <= 9; j++)
            {
                int k = 0;
                while(k < len && strncmp(con, s + k, i + 1) != 0)
                    k ++;
                if(k == len) 
                    break;
                k += i + 1;
                while(k < len && strncmp(con + i + 1, s + k, j - i) != 0)
                    k ++;
                if(k == len) 
                    break;
                k += j - i;
                while(k < len && strncmp(con + j + 1, s + k, 10 - j) != 0)
                    k ++;
                if(k != len)
                {
                    flag = true;
                    break;
                }
            }
        }
        if(flag) 
            puts("YES");
        else 
            puts("NO");
    }
}