NYOJ題目5---Binary String Matching

標籤：

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.

範例輸入

31110011101101011100100100100011010110100010101011 

範例輸出

303 

來源

網路

上傳者

naonao

分析：這道題用strstr函數

#include<iostream>#include<stdio.h>#include<string.h>using namespace std;int main(){    int test,n;    char a[100],b[1000];    cin>>test;    while(test--)    {        cin>>a>>b;        char *p;        p=b;        int count=0;        while(p=strstr(p,a))        {            p=p+1;            count++;        }        printf("%d\n",count);    }    return 0;}

貼一下大牛的最優解

#include<iostream>#include<stdio.h>#include<string.h>using namespace std;int main(){    string a,b;    int n;    cin>>n;    while(n--)    {        cin>>a>>b;        int pos=b.find(a,0);        int  count=0;        while(pos!=string::npos)        {            count++;            pos=b.find(a,pos+1);        }        printf("%d\n",count);    }    return 0;}

NYOJ題目5---Binary String Matching