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 

思路：这题就是就检测，用到for循环。不难

AC代码：

#include<iostream>#include<stdio.h>#include<string.h>using namespace std;int main(){int n,i,j,al,bl,f,s;char a[11],b[1000];while(scanf("%d",&n)!=EOF){while(n--){s=0;scanf("%s%s",a,b);al=strlen(a);bl=strlen(b);            for(i=0;i<bl;i++){if(b[i]==a[0]){   f=0;for(j=1;j<al;j++){if(b[i+j]!=a[j]){f=1;break;}}if(f==0)s++;}}printf("%d\n",s);}}return 0;}

 

标程给我的收获是：1、#include<string>中的find()函数的应用。另外，m!=string::npos  意思是：m不等于字符串的尾部。

标程：

#include<iostream>#include<string>using namespace std;int main(){string s1,s2;int n;cin>>n;while(n--){cin>>s1>>s2;unsigned int m=s2.find(s1,0);int num=0;while(m!=string::npos){num++;m=s2.find(s1,m+1);}cout<<num<<endl;}}