HDU 5311 Sequence

Hidden String

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

Total Submission(s): 803    Accepted Submission(s): 302

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

field=problem&key=BestCoder+1st+Anniversary+%28%24%29&source=1&searchmode=source" style="color:rgb(26,92,200); text-decoration:none">BestCoder 1st Anniversary ($)

 

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#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>

using namespace std;

char str[10001];
struct node
{
    int x;
    char s;
} q[10010];

int main()
{
    int T;
    char a[] = "anniversary";
    scanf("%d",&T);
    while(T--)
    {
        int flag = 0;
        scanf("%s",str);
        int l = strlen(str);
        int ll = strlen(a);
        int num = 0;
        for(int j=0; j<l; j++)
        {
            for(int i=1; i<=ll; i++)
            {
                if(strncmp(str+j,a+num,i) == 0)
                {
                    int num1 = num + i;
                    for(int jj=j+i; jj<l; jj++)
                    {
                        for(int ii=1; ii<=ll-i; ii++)
                        {
                            if(strncmp(str+jj,a+num1,ii) == 0)
                            {
                                int num2 = num1 + ii;
                                for(int jjj=jj+ii; jjj<l; jjj++)
                                {
                                    if(strncmp(str+jjj,a+num2,ll-num2) == 0)
                                    {
                                        printf("YES\n");
                                        flag = 1;
                                        break;
                                    }
                                }
                            }
                            if(flag == 1)
                            {
                                break;
                            }
                        }
                        if(flag == 1)
                        {
                            break;
                        }
                    }
                    if(flag == 1)
                    {
                        break;
                    }
                }
                if(flag == 1)
                {
                    break;
                }
            }
            if(flag == 1)
            {
                break;
            }

        }
        if(flag == 0)
        {
            printf("NO\n");
        }
    }
    return 0;
}