矿大OJ 1768.Power Strings.

题目描述

Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n).

输入

Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.

输出

For each s you should print the largest n such that s = a^n for some string a.

样例输入

abcd
aaaa
ababab
.

样例输出

1
4
3


好像弄成英文看起来就变得高级了 其实就是求emmm.一时间我也不好描述..代码如下

#include <iostream>
#include <string>
using namespace std;
int test(string s)
{
    char a = s[0];
    int locate = s.find(a, 1);
    while(locate > 0)
    {
        bool flag = false;
        int len = s.length();
        int times =  len/ locate;
        if (len % locate == 0)
        {
            flag = true;
            for (int i = locate;i < len&&flag==true;i += locate)
            {
                for (int j = 0;j <= locate - 1;j++)
                {
                    if (s[j] != s[i + j])
                    {
                        flag = false;
                        break;
                    }
                }

            }
            
        }    
        if(flag)
            return times;    
        else
            locate = s.find(a, locate + 1);
    }
    return 1;
}

int main()
{
    string s;
    while (cin >> s && s != ".")
    {
        cout << test(s) << endl;
    }
    return 0;
}

这个做了我差不多5个小时..据说好像用什么KMP一些就做出来，现在还没学也看不懂，等到时候接触了再说吧