紫书第三章练习 uva455 Periodic Strings by crq

A character string is said to have period k if it can be formed by concatenating one or more repetitions of another string of length k. For example, the string "abcabcabcabc" has period 3, since it is formed by 4 repetitions of the string "abc". It also has periods 6 (two repetitions of "abcabc") and 12 (one repetition of "abcabcabcabc").
Write a program to read a character string and determine its smallest period.

Input
The first line oif the input file will contain a single integer N indicating how many test case that your program will test followed by a blank line. Each test case will contain a single character string of up to 80 non-blank characters. Two consecutive input will separated by a blank line.

Output
An integer denoting the smallest period of the input string for each input. Two consecutive output are separated by a blank line.

Sample Input
1

HoHoHo

Sample Output
2

题意：给定一个字符串，它是由某个长度为x的字符串重复若干次产生的，问x的最小值是多少，比如题意中的abcabcabcabc，是由abc重复4次产生的，abc这个字符串的长度为3，所以输出3。

求解：因为字符串长度最大只有80个字符，因此可以直接暴力解决。这里设将字符串分成i段（i从大到小遍历，从而保证x从小到大遍历），由此求出x的值，并判断每段长度为x的子串是否都一样，只要找到一组即退出，便是x的最小值。

AC代码：

#include <iostream>
#include <string>
using namespace std;

int main()
{
    string s;
    int n;
    cin>>n;
    for(int t=0;t<n;t++)
    {
        cin>>s;
        int i, len = s.length();
        int res = -1;
        for(i=len;i>=1;i--)
        {
            if(len%i==0)
            {
                int x = len/i;
                string ss = s.substr(0,x);
                int pos = x, flag = 0;
                while(pos<len)
                {
                    string ss2 = s.substr(pos, x);
                    if(ss2!=ss)
                    {
                        flag = 1;
                        break;
                    }
                    pos += x;
                }
                if(flag==0)
                {
                    res = x;
                    break;                    
                }
            }
        }
        if(res==-1)
            res = len;
        if(t)
            cout<<endl;
        cout<<res<<endl;
    }
    return 0;
}