内存限制:64MB 时间限制:3000ms 特判: No
通过数:232 提交数:458 难度:3
The first line consist only one integer N, indicates N cases follows. In each case, there are two lines, the first line gives the string A, length (A) <= 10, and the second line gives the string B, length (B) <= 1000. And it is guaranteed that B is always longer than A.
For each case, output a single line consist a single integer, tells how many times do B appears as a substring of A.
3 11 1001110110 101 110010010010001 1010 110100010101011
3 0 3
可以用find。
注意用getchar()。。。。。。。
C++代码:
#include<iostream> #include<string> #include<algorithm> #include<cstdio> using namespace std; int main(){ int T; scanf("%d",&T); getchar(); //必须加上 while(T--){ string s1; string s2; getline(cin,s1); getline(cin,s2); int ans = s2.find(s1,0); int num = 0; while(ans >= 0){ num++; ans = s2.find(s1,ans+1); } printf("%d\n",num); } return 0; }
(find) nyoj5-Binary String Matching
原文:https://www.cnblogs.com/Weixu-Liu/p/10532078.html