题目描述
You have a string ss and a chip, which you can place onto any character of this string.
After placing the chip, you move it to the right several (maybe zero) times, i. e. you perform the following operation several times: if the current position of the chip is ii , you move it to the position i + 1i+1 . Of course, moving the chip to the right is impossible if it is already in the last position.
After moving the chip to the right, you move it to the left several (maybe zero) times, i. e. you perform the following operation several times: if the current position of the chip is ii , you move it to the position i - 1i−1 . Of course, moving the chip to the left is impossible if it is already in the first position.
When you place a chip or move it, you write down the character where the chip ends up after your action. For example, if ss is abcdef, you place the chip onto the 33 -rd character, move it to the right 22 times and then move it to the left 33 times, you write down the string cdedcb.
You are given two strings ss and tt . Your task is to determine whether it's possible to perform the described operations with ss so that you write down the string tt as a result.
输入格式
The first line contains one integer qq ( 1 \le q \le 5001≤q≤500 ) — the number of test cases.
Each test case consists of two lines. The first line contains the string ss ( 1 \le |s| \le 5001≤∣s∣≤500 ), the second line contains the string tt ( 1 \le |t| \le 2 \cdot |s| - 11≤∣t∣≤2⋅∣s∣−1 ). Both strings consist of lowercase English characters.
It is guaranteed that the sum of |s|∣s∣ over all test cases does not exceed 500500 .
输出格式
For each test case, print "YES" if you can obtain the string tt by performing the process mentioned in the statement with the string ss , or "NO" if you cannot.
You may print each letter in any case (YES, yes, Yes will all be recognized as positive answer, NO, no and nO will all be recognized as negative answer).
题意翻译
你有两个字符串 s,t 以及一个棋子。接下来你需要进行下列操作：
- 将棋子摆放在字符串 s 的任意一个位置上；
- 将棋子向右移若干次（可以是零次，不能移出字符串）；
- 将棋子向左移若干次（也可以是零次，也不能移出字符串）；
在进行操作的过程中，你需要把棋子经过的字符按顺序写下来。例如， s="abcdef"，你把棋子放在第三个字符上，向右移动两次，再向左移动三次，得到的字符串就是 "cdedcb"。
给定字符串 s,t，你需要求出是否存在一种放置、移动棋子的方案使得最终写下的字符串与 t 相同。有输出 YES，没有输出 NO。 q 组数据。
| 1≤q≤500;1≤∣s∣,∑∣s∣≤500;1≤∣t∣≤2×∣s∣−1;
输入输出样例
输入 #1复制
6 abcdef cdedcb aaa aaaaa aab baaa ab b abcdef abcdef ba baa
输出 #1复制
1. YES 2. YES 3. NO 4. YES 5. YES 6. NO
说明/提示
Consider the examples.
The first test case is described in the statement.
In the second test case, you can place the chip on the 1 -st position, move it twice to the right, and then move it twice to the left.
In the fourth test case, you can place the chip on the 22 -nd position, and then don't move it at all.
In the fifth test case, you can place the chip on the 1 -st position, move it 55 times to the right, and then finish the process.
#include<iostream> #include<cstdio> #include<cstring> using namespace std; const int N=1000+13; char s[N],t[N]; int n,m; inline void solve(){ scanf("%s%s",s+1,t+1);n=strlen(s+1),m=strlen(t+1); for(int i=1;i<=n;++i){ if(s[i]!=t[1]) continue;//如果没有出现和第一位字符相同的就跳过 int r=1; while(i+r<=n&&1+r<=m&&s[i+r]==t[1+r]) ++r;//这里的i是符合条件的 ，在向右加的过程中不要超过n,m； if(r==m) return void(puts("YES"));//比如m=n=1 for(int j=1;j<=r;++j){//这里你可能会认为·直接为j=r，那么你就错了，好好想一想为什么 int l=0; while(i+j-l-2>=1&&l+1+j<=m&&s[i+j-l-2]==t[l+1+j]) ++l;//为什么会减2呢，那是因为我们r的初始值为1哦 if(l==m-j) return void(puts("YES"));//想象一下如果最完美的情况j=r，那么这样是不是就是相当于等长 } } puts("NO"); } int main(){ int T;scanf("%d",&T); while(T--) solve(); return 0; }
