A. Shuffle Hashing-------思维(水)

Polycarp has built his own web service. Being a modern web service it includes login feature. And that always implies password security problems.

Polycarp decided to store the hash of the password, generated by the following algorithm:

take the password p, consisting of lowercase Latin letters, and shuffle the letters randomly in it to obtain p′ (p′ can still be equal to p);
generate two random strings, consisting of lowercase Latin letters, s1 and s2 (any of these strings can be empty);
the resulting hash h=s1+p′+s2, where addition is string concatenation.
For example, let the password p= “abacaba”. Then p′ can be equal to “aabcaab”. Random strings s1= “zyx” and s2= “kjh”. Then h= “zyxaabcaabkjh”.

Note that no letters could be deleted or added to p to obtain p′, only the order could be changed.

Now Polycarp asks you to help him to implement the password check module. Given the password p and the hash h, check that h can be the hash for the password p.

Your program should answer t independent test cases.

Input
The first line contains one integer t (1≤t≤100) — the number of test cases.

The first line of each test case contains a non-empty string p, consisting of lowercase Latin letters. The length of p does not exceed 100.

The second line of each test case contains a non-empty string h, consisting of lowercase Latin letters. The length of h does not exceed 100.

Output
For each test case print the answer to it — “YES” if the given hash h could be obtained from the given password p or “NO” otherwise.

Example
inputCopy
5
abacaba
zyxaabcaabkjh
onetwothree
threetwoone
one
zzonneyy
one
none
twenty
ten
outputCopy
YES
YES
NO
YES
NO
Note
The first test case is explained in the statement.

In the second test case both s1 and s2 are empty and p′= “threetwoone” is p shuffled.

In the third test case the hash could not be obtained from the password.

In the fourth test case s1= “n”, s2 is empty and p′= “one” is p shuffled (even thought it stayed the same).

In the fifth test case the hash could not be obtained from the password.

题意:给你一个p序列,p’为p序列打乱的，h=s1+p’+s2。这里的+都是连接的意思。现在题目问你给你p和h，检查h是否可以得到p。

解析:这道题只要暴力就行了，因为数据范围很小。分三种情况
序列p的长度为n,序列h的长度为m
第一种m<n 肯定输出NO
第二种m==n 检查自身是不是。
第三种 n>m 暴力遍历就行了

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+1000;
int t;
char s[N],str[N];
map<char,int> v,res;
int main()
{
	cin>>t;
	while(t--)
	{
		cin>>(s+1)>>(str+1);
		v.clear();res.clear();
		int n=strlen(s+1);
		int m=strlen(str+1);
		for(int i=1;i<=n;i++) v[s[i]]++;
		if(m<n)
		{
			puts("NO");
			continue;
		}
		if(m==n)
		{
			for(int i=1;i<=m;i++) res[str[i]]++;
			int f=1;
			for(int i=1;i<=m;i++)
			{
				if(v[s[i]]!=res[s[i]])
				{
					f=0;
					break;
				}
			 } 
			if(f==1) 
				puts("YES");
			else puts("NO");
			continue;
		}
		int f=1;
		for(int i=1;i<=m-n+1;i++)
		{
			res.clear();
			for(int j=i;j<=i+n-1;j++) res[str[j]]++;
			 f=1;
			for(int j=1;j<=n;j++)
			{
				if(v[s[j]]!=res[s[j]])
				{
					f=0;
					break;
				}
			}
			if(f==1) 
			{
				puts("YES");
				break;
			}
			    
		}
		if(f==0) puts("NO");
	}
 }