NYOJ 5 Binary String Matching

Binary String Matching

时间限制：3000 ms  |  内存限制：65535 KB

难度：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

输入

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.

输出

For each case, output a single line consist a single integer, tells how many times do B appears as a substring of A.

样例输入

31110011101101011100100100100011010110100010101011 

样例输出

303 

来源

网络

上传者

naonao

/*使用find函数*/#include <iostream>#include <string>using namespace std;int main(){    int t;    string a, b;    cin >> t;    while(t--){        cin >> a >> b;        int n = 0;        int index = b.find(a,0);//返回从0开始找到子串在串中的位置下标        cout<<index;        while(index != b.npos)   //npos表示不存在{           n++;           index = b.find(a, index + 1);        }        cout << n << endl;    }    return 0;}

#include<stdio.h>#include<string.h>int main(){int N,lena,lenb;char a[15],b[1005];scanf("%d",&N);while(N--){int i,j,ans=0;scanf("%s%s",a,b);lena=strlen(a);lenb=strlen(b);for(i=0;i<lenb;i++){int flag=1;for(j=0;j<lena;j++)if(a[j]!=b[i+j]){flag=0;break;}if(flag==1){ans++;}}printf("%d\n",ans);}return 0;}