BestCoder 1st Anniversary ($) 1002 Hidden String

Hidden String

 

 Accepts: 437

 

 Submissions: 2174

 Time Limit: 2000/1000 MS (Java/Others)

 

 Memory Limit: 262144/262144 K (Java/Others)

Problem Description

Today is the 1st anniversary of BestCoder. Soda, the contest manager, gets a string ss of length nn. He wants to find three nonoverlapping substrings s[l_1..r_1]s[l​1​​..r​1​​], s[l_2..r_2]s[l​2​​..r​2​​], s[l_3..r_3]s[l​3​​..r​3​​] that:

1. 1 le l_1 le r_1 < l_2 le r_2 < l_3 le r_3 le n1≤l​1​​≤r​1​​<l​2​​≤r​2​​<l​3​​≤r​3​​≤n

2. The concatenation of s[l_1..r_1]s[l​1​​..r​1​​], s[l_2..r_2]s[l​2​​..r​2​​], s[l_3..r_3]s[l​3​​..r​3​​] is "anniversary".

Input

There are multiple test cases. The first line of input contains an integer TT (1 le T le 100)(1≤T≤100), indicating the number of test cases. For each test case:

There's a line containing a string ss (1 le |s| le 100)(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

#include <stdio.h>
#include <string.h>

char a[105],compare[15]={"anniversary"};
int main()
{
    int T;
    int flg;
    int n,i,j,k,x,y,z,h,h2;
    scanf("%d",&T);
    while(T--)
    {
        flg=0;
        scanf("%s",a);
        n=strlen(a);
        int c=0;
        for(i=0;i<n;i++)
        {
            if(flg==1)
                break;
            if(a[i]==compare[0] && c<=10)
            {
                // printf("hi x%d
",i);
                x=i,c=0;
                while(a[x]==compare[c] && c<=10)
                {
                    // printf("C:%d
",c);
                    if(c==10)
                    {
                        flg=1;
                        break;
                    }
                    c++,x++;
                }
                for(j=x+1;j<n;j++)
                {
                    if(flg==1)
                        break;
                    if(a[j]==compare[c] && c<=10)
                    {
                        h=c;
                        // printf("hi y%d
",j);
                        y=j;
                        while(a[y]==compare[c] && c<=10)
                        {
                            // printf("C:%d
",c);
                            if(c==10)
                            {
                                flg=1;
                                break;
                            }
                            c++,y++;
                        }
                        for(k=y+1;k<n;k++)
                        {
                            if(flg==1)
                                break;
                            if(a[k]==compare[c] && c<=10)
                            {
                                h2=c;
                                // printf("hi z%d
",k);
                                z=k;
                                while(a[z]==compare[c] && c<=10)
                                {
                                    // printf("C:%d
",c);
                                    if(c==10)
                                    {
                                        flg=1;
                                        break;
                                    }
                                    c++,z++;
                                }
                                c=h2;
                            }
                        }
                        c=h;
                    }

                }
            }
        }
        if(flg==1)
            printf("YES
");
        else
            printf("NO
");
    }
    return 0;
}