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NYoj-Binary String Matching-BF算法
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
//******BF算法***********/ #include<cstdio> #include<cstring> #include<iostream> #include<cmath> #include<algorithm> using namespace std; #define MAXSTRLEN 1000 int main() { char s[MAXSTRLEN+10]; char t[11]; int T; scanf("%d",&T); while(T--) { scanf("%s %s",t,s); int j=0,i=0; int len_s=strlen(s); int len_t=strlen(t); int count=0; while(i<=len_s) { if(s[i]==t[j]) { ++i; ++j; if(j==len_t) { count++; i=i-len_t+1; j=0; } } else { i=i-j+1; j=0; } } printf("%d\n",count); } return 0; }
NYoj-Binary String Matching-BF算法
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