编程之美--3.1

题目描述：给定s1,s2两个字符串，要求判定s2是否能够被s1做循环移位得到的字符串包含。

思路：类似于面试金典字符串那一节最后一题，字符串拼接的思路，将s1 = s1+s1，这样只要能够移位得到包含s2那么就肯定在拼接后的s1中

#include <iostream>
#include <string>
#include <fstream>
#include <map>
#include <algorithm>
#include <vector>
#include <ctime>
#include <bitset>
#include <stack>
#include <deque>
#include <stdexcept>
#include <memory>

using namespace std;

int main()
{
    string s1,s2;
    cin>>s1>>s2;
    if(s2.length() > s1.length())
        cout<<"false"<<endl;
    else
    {
        s1 = s1+s1;
        int i;
        for(i = 0 ; i <= s1.length()-s2.length() ; ++i)
        {
            string tmp = s1.substr(i,s2.length());
            cout<<tmp<<" "<<s2<<endl;
            if(tmp == s2)
            {
                cout<<"true"<<endl;
                return 0;
            }
        }
        cout<<"false"<<endl;
    }
    return 0;
}