题目描述:
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
输入描述:
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.
输出描述:
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
题目解析:
本题题意为在一个字符串ss中寻找子串s出现的次数
测试地址:测试点击此处
本题代码:
#include<iostream>
#include<string>
using namespace std;
void deal(string ss,string s)
{
int sum=0,p;
while(1)
{
p=ss.find(s);
if(p==-1)
break;
sum++;
ss.replace(p,1,"q");//利用替换来防止重复查找
}
cout<<sum<<endl;
}
int main()
{
int n;
string s,ss;
cin>>n;
while(n--)
{
cin>>s>>ss;
deal(ss,s);
}
return 0;
}