一、題目描述

 二、解題思路

　　簡單做法用c++中的stl中string類的find函式一句話解決

三、程式碼實現

 1 #include "bits/stdc++.h"
 2 #define PII pair<int,int>
 3 #define rep(i,z,n) for(int i = z;i <= n; i++)
 4 #define per(i,n,z) for(int i = n;i >= z; i--)
 5 #define ll long long
 6 #define db double
 7 #define vi vector<int>
 8
 
 #define debug(x) cerr << "!!!" << x << endl;
 9 using namespace std;
10 //從某個串中把某個子串替換成另一個子串
11 string& replace_all(string& src, const string& old_value, const string& new_value) {
12     // 每次重新定位起始位置，防止上輪替換後的字串形成新的old_value
13     for (string::size_type pos(0); pos != string
 
::npos; pos += new_value.length()) {
14         if ((pos = src.find(old_value, pos)) != string::npos) {
15             src.replace(pos, old_value.length(), new_value);
16         }
17         else break;
18     }
19     return src;
20 }
21 inline ll read()
22 {
23     ll s,r;
24     r = 1;
25     s = 0
 
;
26     char ch = getchar();
27     while(ch < '0' || ch > '9'){
28         if(ch == '-')
29             r = -1;
30         ch = getchar();
31     }
32     while(ch >= '0' && ch <= '9'){
33         s = (s << 1) + (s << 3) + (ch ^ 48);
34         ch = getchar();
35     }
36     return s * r;
37 }
38 inline void write(ll x)
39 {
40     if(x < 0) putchar('-'),x = -x;
41     if(x > 9) write(x / 10);
42     putchar(x % 10 + '0');
43 }
44 int main()
45 {
46     string a,b;
47     int n;
48     n = read();
49     while(n--){
50         cin >> a >> b;
51         if(a.find(b) != 0)
52             cout << "Yes" << endl;
53         else 
54             cout << "No" << endl;
55     }
56     return 0;
57 }