NYOJ 5 Binary String Matching

最新推荐文章于 2018-05-03 21:51:44 发布

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本文介绍了一种解决二进制字符串A作为子串在字符串B中出现次数的问题。通过C++实现，利用字符串操作如substr和compare来查找并计数。文章提供了完整的代码示例及运行方式。

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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

3
11
1001110110
101
110010010010001
1010
110100010101011

样例输出

3
0
3

AC代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
string s,s1,s2;
cin>>s1>>s2;
int l1,l2;
l1=s1.length();//C++不用这种格式：strlen(s1)
l2=s2.length();
int ans=0;
for(int i=0;i<l2-l1+1;i++)
{
s=s2.substr(i,l1);
//substr() 方法可在字符串中抽取从i下标开始的指定数目l的字符
if(s.compare(s1)==0)
//上述 s.compare(s1) 是C++中比较两个字符串是否相同
//不用C这种格式：strcmp(s1，s2) 判断两个字符串s1和s2是否相同
ans++;
}
printf("%d\n",ans);
}
return 0;
}