Binary String Matching
时间限制：
3000 ms | 内存限制：
65535 KB
难度：
3
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描述
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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
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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
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3 0 3
来源
- 网络 上传者
naonao
代码
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<stack>
#include<queue>
#define inf 0x3f3f3f
#define M 10000+10
using namespace std;
char str[M];
char ss[20];
int main()
{
int n;
scanf("%d",&n);
while(n--)
{
scanf("%s",ss);
scanf("%s",str);
int sum=0;int j;
for(int i=0;i<strlen(str);i++)
{
int k=i;
for(j=0;j<strlen(ss);j++) // 一个个比对
{
if(ss[j]!=str[k++])
{
break;
}
}
if(j==strlen(ss))
sum++;
}
printf("%d\n",sum);
}
return 0;
}