TZOJ 2406 Power Strings（KMP找最多重复子串）

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

提示

This problem has huge input, use scanf instead of cin to avoid time limit exceed.

思路：

就是问一个字符串写成(a)^n的形式，求最大的n.

根据KMP的next函数的性质，已知字符串t第k个字符的next[k],那么d=k-next[k],如果k%d==0,那么t[1……k]最多可均匀的分成k/d份。也就是可以生成一个长度为d的重复度为k/d的字串。

#include<bits/stdc++.h>
using namespace std;
const int M=1e6+5;
char t[M];
int next[M],tlen;
void getNext()
{
    int i=0,j=-1;
    next[0]=-1;
    while(i<tlen)
    {
        if(j==-1||t[i]==t[j])
            next[++i]=++j;
        else j=next[j];
    }
}
int main()
{
    while(scanf("%s",t)!=EOF,t[0]!='.')
    {
        tlen=strlen(t);
        getNext();
        if(tlen%(tlen-next[tlen])==0)
            printf("%d
",tlen/(tlen-next[tlen]));
        else printf("1
");
    }
    return 0;
}