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
View Code#include<stdio.h> #include<string.h> #define maxn 1000+10 char strb[maxn]; char stra[11]; int main() { int n,i,count,j,lena,lenb,k; scanf("%d",&n); getchar(); while(n--) { scanf("%s",stra); scanf("%s",strb); lena=strlen(stra); lenb=strlen(strb); count=0;j=0; for(i=0;i<lenb;i++) {j=0;k=i; while(1&&j<lena) { if(stra[j]!=strb[k]) break; j++;k++; } if(j==lena) count++; } printf("%d\n",count); } return 0; }