Binary String Matching【暴力枚举】【字符串的处理】

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 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;  }