[POJ 2406]Power Strings（kmp）

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Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 50554		Accepted: 21087

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

Hint

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

Source

Waterloo local 2002.07.01

題意：就是問一個字符串寫成(a)^n的形式，求最大的n.

題解：根據KMP的next函數的性質，已知字符串t第K個字符的next[k],那麼d=k-next[k],如果k%d==0,那麼t[1……k]最多可均勻的分成k/d份。也就是可以生成一個長度爲d的重複度爲k/d的字串。

代碼：

#include<iostream>
#include<cstring>
#include<stdio.h>
using namespace std;

const int MAX=1e6+10;
char s[MAX];
char t[MAX];
int Next[MAX]; 
int n,m;
int ans=0;

void getNext(){
	int i=0,j=0;
	Next[0]=-1;
	j=Next[i];
	while(i<m){
		if(j==-1||t[i]==t[j]){
			i++;
			j++;
			Next[i]=t[i]==t[j]?Next[j]:j;
		}
		else{
			j=Next[j];
		}
	}
}

int main(){
	while(~scanf("%s",t)){
		if(t[0]=='.')
		break;
		m=strlen(t);
		getNext();
		ans=1;//最小是1，不初始化會WA
		if(m%(m-Next[m])==0)//判斷是否能整除

		ans=m/(m-Next[m]);

		cout<<ans<<endl;
	}
	return 0;
}