HDU5311——字符串——Hidden String

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 ($)

 

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/*
先求出前缀和后缀匹配，注意anniversaxxrxxy这组数据。。被卡了一上午，，因为访问到第二个a的时候会让前缀变成1
然后就for循环找中间匹配的串，保证这个串的第一个和最后一个与头尾相连
比赛的时候狗血代码都过了- -。。。如果自己hack自己涨分的话......
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;

char  s1[110];
char s2[110] = "anniversary";
int pre1[110], pre2[110];
int T;
int main()
{
    scanf("%d", &T);
    while(T--){
        scanf("%s", s1);
        int len1 = strlen(s1);
        int len2 = strlen(s2);
        memset(pre1, 0, sizeof(pre1));
        memset(pre2, 0 ,sizeof(pre2));
        int cout = 0;
        for(int i = 0 ; i < len1; i++){
            if(s1[i] == 'a' && cout == 0){
                pre1[i] = 1;
                cout = 1;
            }
            else if(s1[i] == s2[cout]){
                pre1[i] = pre1[i-1] + 1;
                cout++;
            }
            else {
                pre1[i] = 0;
                cout= 0;
            }
        }
        //    for(int i = 0; i < len1; i++)
        //        printf("%d ",pre1[i]);
        cout = 0;
        for(int i = len1 - 1; i >= 0 ; i--){
            if(s1[i] == 'y'){
                pre2[i] = 1;
                cout = 1;
            }
            else if(s1[i] == s2[len2 - cout-1]){
                pre2[i] = pre2[i+1] + 1;
                cout++;
            }
            else {
                pre2[i] = 0;
                cout = 0;
            }
        }
        //  for(int i = 0 ; i < len1; i++)
        //      printf("%d ",pre2[i]);

        int flag = 0;

        cout = 0;
        for(int i = 0 ; i < len1 ; i++){
            for(int j = i + 1; j < len1; j++){
                cout = 0;
                for(int k = i + 1; k < j; k++){
                    if(s1[k] == s2[pre1[i]+cout]){
                        cout++;
                    }
                    else {
                        cout = 0;
                        if(s1[k] == s2[pre1[i]+cout])
                            cout++;
                    }
                    if(cout + pre1[i] + pre2[j] == len2 && s1[k] == s2[len2 - pre2[j] - 1]){
                        flag = 1;
                        break;
                    }
                }
                if(flag) break;
            }
            if(flag) break;
        }
        //   for(int i = 0 ; i < len1; i++)
        if(flag) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}