NYOJ 5 Binary String Mathing (substr函式)
阿新 • • 發佈:2019-01-06
Binary String Matching
時間限制:3000 ms | 記憶體限制:65535 KB 難度:3- 描述
- Given two strings A and B, whose alphabet consist only ‘0’ and ‘1’. Your task is only to tell how many times does A appear as a substring of B? For example, the text string B is ‘1001110110’ while the pattern string A is ‘11’, you should output 3, because
the pattern A appeared at the posit
- 輸入
- 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.
- 輸出
- For each case, output a single line consist a single integer, tells how many times do B appears as a substring of A.
- 樣例輸入
-
3 11 1001110110 101 110010010010001 1010 110100010101011
- 樣例輸出
-
3 0 3
-
#include<stdio.h> #include<iostream> #include<algorithm> #include<string.h> using namespace std; int main() { int T; string a; string b; cin>>T; while(T--) { cin>>a; cin>>b; int count=0; int len1=a.size(); int len2=b.size(); for(int i=0;i<len2;i++) { if(b.substr(i,len1)==a.substr(0,len1)) count++; } cout<<count<<endl; } return 0; }