Binary String Matching
时间限制：
3000 ms | 内存限制：
65535 KB
难度：
3
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描述
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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
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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
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3 11 1001110110 101 110010010010001 1010 110100010101011
样例输出
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3 0 3
来源
- 网络 上传者
naonao
分析：这道题用strstr函数
#include<iostream> #include<stdio.h> #include<string.h> using namespace std; int main() {int test,n;char a[100],b[1000];cin>>test;while(test--){cin>>a>>b;char *p;p=b;int count=0;while(p=strstr(p,a)){p=p+1;count++;}printf("%d\n",count);}return 0; }
贴一下大牛的最优解#include<iostream> #include<stdio.h> #include<string.h> using namespace std; int main() {string a,b;int n;cin>>n;while(n--){cin>>a>>b;int pos=b.find(a,0);int count=0;while(pos!=string::npos){count++;pos=b.find(a,pos+1);}printf("%d\n",count);}return 0;}