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POJ 2406 Power Strings

2024-07-19 10:54:54   223人阅读

Description

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).

Input

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.

Output

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

Sample Input

abcd
aaaa
ababab
.

Sample Output

1
4
3

题意：给你一个字符串，要你找到一个最大的n使得a^n=该字符串（a为该字符串的子串）。
思路：既然是找最大的n，那么我只要找该字符串的最小周期，那不正好可以解决这个问题了？依照这个思路来敲代码！
所以AC代码：
#include<stdio.h>
#include<string.h>
#define N 1000005
char str[N];
int main()
{
    int sum;
    while(scanf("%s",str)!=EOF)
    {
        int len=strlen(str);
        if(str[0]=='.')break;
        for(int i=1;i<=len;i++)     //一个剪枝，如果i不能被Len整除，那么i绝对不是字符的周期子串长
            if(len%i==0)
            {
                int ok=1;
                for(int j=i;j<len;j++)
                    if(str[j]!=str[j%i]){ok=0;break;}
                if(ok)
                {
                    sum=i;
                    break;
                }
            }
        printf("%d\n",len/sum);
    }
    return 0;
}


POJ 2406 Power Strings


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