原题如下

Polycarp has a string s[1…n] of length n consisting of decimal digits. Polycarp performs the following operation with the string s no more than once (i.e. he can perform operation 0 or 1 time):

Polycarp selects two numbers i and j (1≤i≤j≤n) and removes characters from the s string at the positions i,i+1,i+2,…,j (i.e. removes substring s[i…j]). More formally, Polycarp turns the string s into the string s1s2…si−1sj+1sj+2…sn.
For example, the string s=“20192020” Polycarp can turn into strings:

“2020” (in this case (i,j)=(3,6) or (i,j)=(1,4));
“2019220” (in this case (i,j)=(6,6));
“020” (in this case (i,j)=(1,5));
other operations are also possible, only a few of them are listed above.
Polycarp likes the string “2020” very much, so he is wondering if it is possible to turn the string s into a string “2020” in no more than one operation? Note that you can perform zero operations.

Input
The first line contains a positive integer t (1≤t≤1000) — number of test cases in the test. Then t test cases follow.

The first line of each test case contains an integer n (4≤n≤200) — length of the string s. The next line contains a string s of length n consisting of decimal digits. It is allowed that the string s starts with digit 0.

Output
For each test case, output on a separate line:

“YES” if Polycarp can turn the string s into a string “2020” in no more than one operation (i.e. he can perform 0 or 1 operation);
“NO” otherwise.
You may print every letter of “YES” and “NO” in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).

input

6
8
20192020
8
22019020
4
2020
5
20002
6
729040
6
200200

output
YES
YES
YES
NO
NO
NO

题目大意

给你一个字符串，问是否能进行不超过一次的操作（删除从i至j的字符）是字符串变为“2020”，如果可以输出“YES”，否则输出"NO"。

由于只能进行一次或者0次操作，说明该字符串的首尾必须符合一下六种类型：

• 2020
• 2020XXXXX
• 2XXXXX020
• 20XXXXX20
• 202XXXXX0
• XXXXX2020

在知道这么个规律之后，我们便可直接用if语句解决这个问题

代码如下:

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
const int N = 1e4+10;

using namespace std;

int main(){
    
	int n;
	cin >> n;
	for(int i = 0;i < n;i++){
    
		char a[N];
		int x;
		cin >> x;   //字符串的长度
		for(int j = 0;j < x;j++) cin >> a[j];   //输入字符串
		if(a[x-1]=='0'&&a[x-2]=='2'&&a[x-3]=='0'&&a[x-4]=='2')
		 cout << "YES" << endl;
		else if(a[x-1] == '0'&&a[x-2]=='2'&&a[x-3]=='0'&&a[0]=='2')
		 cout << "YES" << endl;
		else if(a[x-1] == '0'&&a[x-2]=='2'&&a[1]=='0'&&a[0]=='2')
		 cout << "YES" << endl;
		else if(a[x-1] == '0'&&a[2]=='2'&&a[1]=='0'&&a[0]=='2')
		 cout << "YES" << endl;
		else if(a[3] == '0'&&a[2]=='2'&&a[1]=='0'&&a[0]=='2')
		 cout << "YES" << endl;
		else cout << "NO" << endl;
	}
	return 0;
	}

看懂了吗，看不懂也得懂（doge）

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