POJ 2406 Power Strings（kmp）

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原创最新推荐文章于 2026-07-30 10:09:02 发布 1k 阅读 AI 写同款· GEO 优化 ›

数学建模种群竞争模型学习笔记模型背景应用场景种群竞争模型例如 s1=0.5，则表示对于供养甲的资源来讲，单位数量的乙的消耗为单位数量的甲的0.5倍，即表示乙对于该资源的占有能力是甲的0.5倍（即竞争力弱）仿真结果及分析（1）设置自然增长率 r1=r2=1 （2）改变自然增长率 r1=r2=0.3 （3）修改甲和乙的最大容量 n1=10000, n2=100 （4）修改甲和乙的初始数量 x1=10, x... 阅读详情

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Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 33335		Accepted: 13852

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

求一个串的最小周期

对于一个串，如果abcdabc 那么next[len]=3,那么len-next【len】就大于len/2,那么len%(len-next[len])!=0;而对于一个周期串ababab next[len]=4,此时len-next[len]应该等于

最小串的长度，所以是不是有最小周期就可以用len%(len-next[len])是否为0来判断，（个人理解，如有错，请想告）

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
using namespace std;
#define N 100005

char a[N];
int len,next[N];

void getfail(char *a)
{
	 int i,j;
	 len=strlen(a);
	 i=0;j=-1;
	 next[0]=-1;
	 while(i<len)
	 {
	 	if(j==-1||a[i]==a[j])
		{
			 i++;
			 j++;
			 next[i]=j;
		}
		else
			j=next[j];
	 }
}

int main()
{
	int i,j;
	while(scanf("%s",a))
	{
		if(a[0]=='.') break;
		getfail(a);
		int ans=len%(len-next[len]);
		if(ans==0)
			printf("%d\n",len/(len-next[len]));
		else
			printf("1\n");
	}
	return 0;
}