南阳oj 5 Binary String Matching
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
#include<stdio.h> #include<string.h> int main() { char a[120],b[1200]; int ai,bi; int t,i,j,k,cut; scanf("%d",&t); while(t--){ cut=0; scanf("%s%s",&a,b); ai=a[0]-'0'; int len1=strlen(a); int len2=strlen(b); for(i=1;i<len1;i++){ ai*=10; ai+=a[i]-'0'; } for(i=0;i<len2;i++) { bi=b[i]-'0'; for(k=i+1,j=0;j<len1-1;j++,k++) { bi*=10; bi+=b[k]-'0'; } if(ai==bi) cut++; } printf("%d\n",cut); } return 0; }
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