1、题目信息
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
2、题目大致意思
第一次输入一个整数n,代表此时组数,接着输出n组测试用例,每组包含两行a、b字符串,a的长度<=10,b的长度<=1000
输出b中包含多少个a。
3、思路
1)使用c++ 中string类来处理
2)对于遍历字符串b,每次取出长度为a的字串,进行判断
4、代码
#include <iostream> #include <string> using namespace std; int main() { string a,b; int n,m; cin>>n; while(n--){ m=0; cin>>a; cin>>b; for(int i=0;i<b.length();i++){ if(b.substr(i,a.length())==a) m++; } cout<<m<<endl; } return 0; }