题目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 

我的解题程序

C++:

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

JAVA:

import java.util.Scanner;public class Main {public static void main(String args[]){String str1,str2;int n,num,index;Scanner scanner = new Scanner(System.in);n = scanner.nextInt();while(n-- > 0){str1 = scanner.next();str2 = scanner.next();num = 0;index = str2.indexOf(str1);while(index != -1){num++;index = str2.indexOf(str1, index+1);}System.out.println(num);}}}

程序分析：

     先分析算法思路，在两个程序中都使用了String类和其中提供的方法。只要注意类中方法的使用就可以了。

     也可以自己写函数实现匹配计数功能，容易出现的错误：如str1为“11”,str2为“1110”，容易计数为1，正确应为2。注意算法的逻辑。