杭电5311-Hidden String（暴力列举）

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原创最新推荐文章于 2026-09-25 11:18:11 发布 676 阅读 AI 写同款· GEO 优化 ›

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Hidden String

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

Total Submission(s): 1688 Accepted Submission(s): 595

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

(1≤T≤100), indicating the number of test cases. For each test case:

There's a line containing a string

(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

暴力列举每一段的长度，把目标字符串分为三段。用strstr函数找到第一次出现字符串的地址

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char aim[]="anniversary",s[110],s1[110];
int main()
{
    int i,j,k,m,flag,n,len,l,loca;
    len=strlen(aim);
    scanf("%d",&m);
    while(m--)
    {
        scanf("%s",s);
        flag=0;
        for(i=0;i<=8;i++)
        {
            for(j=i+1;j<=9;j++)
            {
                strcpy(s1,aim);
                s1[i+1]='\0';
                l=strstr(s,s1)-s;
                if(l<0)
                continue;
                l+=i+1;
                
                strcpy(s1,aim+i+1);
                s1[j-i]='\0';
                l=strstr(s+l,s1)-s;
                if(l<0)
                continue;
                l+=j-i;
                
                
                strcpy(s1,aim+j+1);
                l=strstr(s+l,s1)-s;
                if(l<0)
                continue;
                
                flag=1;
                break;
            }
            if(flag)
            break;
        }
        if(flag)
        printf("YES\n");
        else
        printf("NO\n");
    }
    return 0;
 }

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