KMP(3)--poj2406(next数组的应用)

最新推荐文章于 2021-07-31 08:54:13 发布

原创最新推荐文章于 2021-07-31 08:54:13 发布 · 384 阅读

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#poj2406 #KMP #字符串周期

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本文介绍了一种高效算法来解决字符串的周期性问题，即找出一个字符串能被表示为某个子串重复多次的最大次数。通过使用KMP算法的next数组特性，文章提供了一个简洁的C语言实现示例。

Power Strings

Time Limit:3000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu

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.

这道题就是求字符串的周期，即问一个字符串写成(a)^n的形式，求最大的n.
因为字符串可以写成(a)^n的形式，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。接下来就简单了，上代码：

#include<stdio.h>
#include<string.h>
#define MAX 1000010
char str[MAX];
int next[MAX];
void GetNextval(){
	int i=0;
	int j=-1;
	next[0]=-1;
	int len=strlen(str);
	while(i<len){
		if(j==-1||str[i]==str[j]){
			next[++i]=++j;
		}
		else j=next[j];
	}
}
int main()
{
	int i;
	while(scanf("%s",str)!=EOF){
		if(str[0]=='.')break;
		GetNextval();        
		int max=1;
		int len=strlen(str);                 //利用求得的next[len]，按我说的性质就可求得周期
			if(len%(len-next[len])==0)
				max=len/(len-next[len]);
		printf("%d\n",max);
	}
	return 0;
}