CF1451B-Non-Substring Subsequence

题目:

题目描述:

Hr0d1y has $q$ queries on a binary string $s$ of length $n$ . A binary string is a string containing only characters ‘0’ and ‘1’.

A query is described by a pair of integers $l_i$ , $r_i$ $(1 \leq l_i \lt r_i \leq n)$ .

For each query, he has to determine whether there exists a good subsequence in $s$ that is equal to the substring $s[l_i\ldots r_i]$ .

• A substring $s[i\ldots j]$ of a string $s$ is the string formed by characters $s_i s_{i+1} \ldots s_j$ .
• String $a$ is said to be a subsequence of string $b$ if $a$ can be obtained from $b$ by deleting some characters without changing the order of the remaining characters.
• A subsequence is said to be good if it is not contiguous and has length $\ge 2$ . For example, if $s$ is “1100110”, then the subsequences $s_1s_2s_4$ (“1100110”) and $s_1s_5s_7$ (“1100110”) are good, while $s_1s_2s_3$ (“1100110”) is not good.

Can you help Hr0d1y answer each query?

输入格式:

The first line of the input contains a single integer $t$ ( $1\leq t \leq 100$ ) — the number of test cases. The description of each test case is as follows.

The first line contains two integers $n$ ( $2 \leq n \leq 100$ ) and $q$ ( $1\leq q \leq 100$ ) — the length of the string and the number of queries.

The second line contains the string $s$ .

The $i$ -th of the next $q$ lines contains two integers $l_i$ and $r_i$ ( $1 \leq l_i \lt r_i \leq n$ ).

输出格式:

For each test case, output $q$ lines. The $i$ -th line of the output of each test case should contain “YES” if there exists a good subsequence equal to the substring $s[l_i…r_i]$ , and “NO” otherwise.

You may print each letter in any case (upper or lower).

样例:

样例输入1:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10

2
6 3
001000
2 4
1 3
3 5
4 2
1111
1 4
2 3

样例输出1:

1
2
3
4
5

YES
NO
YES
NO
YES

思路:

实现:

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#include "ybwhead/ios.h"
string s;
int n,q;
int main()
{
	int TTT;
	yin>>TTT;
	while(TTT--)
	{
		yin>>n>>q;
		yin>>s;
		for(int i=1;i<=q;i++)
		{
			int l,r;
			yin>>l>>r;--l;--r;
			bool flg=0;
			for(int j=0;j<l;j++)
			{
				if(s[j]==s[l])
				{
					flg=1;break;
				}
			}
			for(int j=r+1;j<n;j++)
			{
				if(s[j]==s[r])
				{
					flg=1;break;
				}
			}
			if(flg)puts("YES");
			else puts("NO");
		}
	}
	return 0;
}