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
1 #include <stdio.h> 2 #include <string.h> 3 4 int main(){ 5 int T; 6 char a[11]; 7 char temp[11]; 8 char b[1001]; 9 int a_length; 10 int b_length; 11 int i; 12 int amount; 13 14 scanf("%d",&T); 15 16 while(T--){ 17 scanf("%s%s",a,b); 18 a_length=strlen(a); 19 b_length=strlen(b); 20 21 amount=0; 22 for(i=0;i<b_length-a_length+1;i++){ 23 memset(temp,0,11); 24 strncpy(temp,&b[i],a_length); 25 26 if(strcmp(temp,a)==0){ 27 amount++; 28 } 29 } 30 printf("%d ",amount); 31 } 32 return 0; 33 }