题干：
Give a string SS and NN string T_iTi , determine whether T_iTi is a subsequence of SS.
If ti is subsequence of SS, print YES,else print NO.
If there is an array \lbrace K_1, K_2, K_3,\cdots, K_m \rbrace{K1,K2,K3,⋯,Km} so that 1 \le K_1 < K_2 < K_3 < \cdots < K_m \le N1≤K1<K2<K3<⋯<Km≤N and S_{k_i} = T_iSki=Ti, (1 \le i \le m)(1≤i≤m), then T_iTi is a subsequence of SS.
Input
The first line is one string SS,length(SS) \le 100000≤100000
The second line is one positive integer N,N \le 100000N,N≤100000
Then next nn lines,every line is a string T_iTi, length(T_iTi) \le 1000≤1000
Output
Print NN lines. If the ii-th T_iTi is subsequence of SS, print YES, else print NO.
样例输入复制
abcdefg 3 abc adg cba
样例输出复制
YES YES NO
题目大意：
给你一个模板串，再给你n个串，问你这n个串能否由模板串的子序列构成。
解题报告：
出过好几次的套路了。。。刚开始NO没加\n，，，wa一发真可惜、、
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
const int MAX = 2e5 + 5;
string ss;
char s[MAX];
int trie[MAX][66];
int main() {
int n;
cin>>ss;
int len = ss.size();
ss = ' ' + ss;
for(int j = 0; j<26; j++) trie[len+1][j] = -1;
for(int i = len; i>=1; i--) {
for(int j = 0; j<26; j++) {
if(j == ss[i] - 'a') trie[i][j] = i;
else trie[i][j] = trie[i+1][j];
}
}
cin>>n;
for(int i = 1; i<=n; i++) {
int cur = 0;
scanf("%s",s);
int up = strlen(s);
for(int j = 0; j<up; j++) {
cur = trie[cur+1][s[j] - 'a'];
if(cur == -1) {
break;
}
}
if(cur != -1) printf("YES\n");
else printf("NO\n");
}
return 0 ;
}