HDU 5311:Hidden String【字符串】

最新推荐文章于 2021-03-04 10:48:50 发布

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最新推荐文章于 2021-03-04 10:48:50 发布

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分类专栏： ACM成长脚印文章标签：字符串 HDU string

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, 本文介绍了一个编程问题，任务是在给定字符串中找到三个不重叠的子串，使其拼接后形成anniversary。文章提供了AC代码示例，通过枚举每段子串的长度并检查是否符合条件。

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Hidden String

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1738 Accepted Submission(s): 612

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，把字符串anniversary分成三段，枚举每段的长度，分析是否有符合条件的。。。

AC-code:

#include<cstdio>
#include<cstring>
int main()
{
	int T,i,j,flag,l;
	char s[105],str[]="anniversary";
	scanf("%d",&T);
	getchar();
	while(T--)
	{
		gets(s);
		l=strlen(s);
		int flag=0;
		for(i=0;i<=8;i++)
		{
			for(j=i+1;j<=9;j++)
			{
				int k=0;
				while(k<l&&strncmp(s+k,str,i+1)!=0)
					k++;
				if(k==l)
					break;
				k+=i+1;
				while(k<l&&strncmp(s+k,str+i+1,j-i)!=0)
					k++;
				if(k==l)
					break;
				k+=j-i;
				while(k<l&&strncmp(s+k,str+j+1,10-j)!=0)
					k++;
				if(k!=l)
				{
					flag=1;
					break;
				}
			}
			if(flag)
				break;
		}
		if(flag)
			puts("YES");
		else 
			puts("NO");
	}
	return 0;
}