poj 2406 Power Strings

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最新推荐文章于 2022-07-21 20:55:14 发布

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本文详细介绍了如何使用KMP算法求解最小重复子串问题，并提供了代码实现及实例解析。

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Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 33121		Accepted: 13777

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.

题意：求最小重复子串；

解题思路：利用KMP算法的性质，用next数组来存储，（next数组的使用：刚开始从第二个字符（后面用j表示）找与第一个（后面用i表示）字符相同的字符，如果不同则令其next为0，找到相同的时候，i++，j++如果相同（即s[i]==s[j]）将next[i]=j；一直执行这个操作（找到相同的时候。。。。将next[i]=j）直到不相同时，j=next[j]，继续执行下去；

#include <iostream>
#include <string.h>
using namespace std;
char s[1000005];
int next[1000005]; 
int len;
void getnext(){	
	len=strlen(s);
	int i=0,j=-1;
	next[0]=-1;
	while (i<len){
		if (j==-1||s[i]==s[j])
			next[++i]=++j;
		else
			j=next[j];
	}
}
int main(){
	while (cin>>s){
		if (strcmp(s,".")==0)
			break;
		len=strlen(s);
		getnext();
		if (len%(len-next[len])==0)
			cout<<len/(len-next[len])<<endl;
		else
			cout<<1<<endl;
	}
	return 0;
}