Cyclic Shift（字符串匹配水题）

最新推荐文章于 2022-07-29 17:47:33 发布

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探讨在给定两个相同长度的字符串情况下，通过一次循环移位子序列使两字符串相等的可能性。代码实现使用C++，遍历并比较字符，判断是否能通过特定操作达到目标。

You are given two strings a and b of the same length and consisting of lowercase English letters. You can pick at most one subsequence of string b and do a cyclic shift on that subsequence exactly once.

For example, if you have a string “abcdefg” and you picked the letters at indices 2, 5, and 6 as a subsequence to do a cyclic shift on them, the letter at index 2 will go to index 5, the letter at index 5 will go to index 6, the letter at index 6 will go to index 2, and the string will become “afcdbeg”.

Your task is to check if it is possible to make string b equivalent to string a using at most one cyclic shift. Can you?
Input

The first line contains an integer T (1 ≤ T ≤ 200) specifying the number of test cases.

The first line of each test case contains an integer n (1 ≤ n ≤ 105) specifying the length of strings a and b. Then two lines follow, giving strings a and b, respectively. Both strings consist only of lowercase English letters.
Output

For each test case, print a single line containing “YES” (without quotes) if it is possible to make string b equivalent to string a using at most one cyclic shift. Otherwise, print “NO” (without quotes).
Example
Input

2
6
abcdef
adcebf
4
abcd
dabd

Output

YES
NO

#include<cstdio>
#include<iostream>
#include<cstring> 
using namespace std;
const int maxn=1e5+5;
char a[maxn],b[maxn],aa[maxn],bb[maxn];
int main(){
	int T;
	scanf("%d",&T);
	while(T--){
		int n;
		scanf("%d",&n);
		scanf("%s",a+1);
		scanf("%s",b+1);
		int t=0;
		for(int i=1;i<=n;i++){
			if(a[i]!=b[i]){
				aa[t]=a[i];
				bb[t]=b[i];
				t++;
			}
		}
		bool flag=true;
		for(int i=0;i<t;i++){
			if(bb[i]==aa[(i+1)%t])
			continue;
			else{
				flag=false;
				break;
			}
		}
		if(flag) printf("YES\n");
		else printf("NO\n");
	}
	return 0;
}