问题 I: From S To T-----最暴力的方法!!!用题目意思直接做！！！！
题目描述
You are given three strings s, t and p consisting of lowercase Latin letters. You may perform any number (possibly, zero) operations on these strings.
During each operation you choose any character from p, erase it from p and insert it into string s (you may insert this character anywhere you want: in the beginning of s, in the end or between any two consecutive characters).
For example, if p is aba, and s is de, then the following outcomes are possible (the character we erase from p and insert into s is highlighted):
aba → ba, de → ade;
aba → ba, de → dae;
aba → ba, de → dea;
aba → aa, de → bde;
aba → aa, de → dbe;
aba → aa, de → deb;
aba → ab, de → ade;
aba → ab, de → dae;
aba → ab, de → dea;
Your goal is to perform several (maybe zero) operations so that s becomes equal to t. Please determine whether it is possible.
Note that you have to answer q independent queries.
输入
The first line contains one integer q (1≤q≤100) — the number of queries. Each query is represented by three consecutive lines.
The first line of each query contains the string s (1≤|s|≤100) consisting of lowercase Latin letters.
The second line of each query contains the string t (1≤|t|≤100) consisting of lowercase Latin letters.
The third line of each query contains the string p (1≤|p|≤100) consisting of lowercase Latin letters.
输出
For each query print YES if it is possible to make s equal to t, and NO otherwise.
样例输入
4
ab
acxb
cax
a
aaaa
aaabbcc
a
aaaa
aabbcc
ab
baaa
aaaaa
样例输入
YES
YES
NO
NO提示
In the first test case there is the following sequence of operation:
s= ab, t= acxb, p= cax;
s= acb, t= acxb, p= ax;
s= acxb, t= acxb, p= a.
In the second test case there is the following sequence of operation:
s= a, t= aaaa, p= aaabbcc;
s= aa, t= aaaa, p= aabbcc;
s= aaa, t= aaaa, p= abbcc;
s= aaaa, t= aaaa, p= bbcc.
一个字符串题目，本来没有想法，在涵大佬的提示下，得到了灵感，get到了自己的暴力枚举法hhhhh
思路都在代码注释
#include
#include
#include
#include
using namespace std;
char s[109],t[109],p[109];
int n,lens,lent,lenp,cnt1=0,cnt2=0;
bool vist[109];
bool visp[109];
int main()
{
scanf("%d",&n);
while(n--)
{
cnt1=0;
cnt2=0;
memset(vist,0,sizeof(vist));
memset(visp,0,sizeof(visp));
scanf("%s",s);
scanf("%s",t);
scanf("%s",p); //思路：先从字符串t里面从前往后按顺序找s里的字符，若找到的匹配的个数与字符串s的长度不符，
lens=strlen(s); //则说明s不是t一个子串，即存在字符属于s但是不属于t，即s和p怎么操作都有不匹配的地方
lent=strlen(t); //若匹配到的字符个数与s长度相同，则说明s是t的一个子串，那么再从p里面匹配t中没有被标记过的字符，
lenp=strlen(p); //查找可以没有顺序，因为可以随机插入，但是找到一个必须要在两个字符串中都要标记 ,最后两个被标记的和如果等于t的长度，则满足要求
int temp=0;
for(int i=0;i