(POJ - 2406)Power Strings

最新推荐文章于 2022-07-21 20:55:14 发布

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最新推荐文章于 2022-07-21 20:55:14 发布

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分类专栏： kmp 文章标签： poj

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本文介绍了解决 POJ-2406 PowerStrings 问题的方法，该问题是寻找给定字符串的最大重复次数。通过使用 KMP 算法求取最短循环节来高效解决这一挑战。

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(POJ - 2406)Power Strings

Time Limit: 3000MS Memory Limit: 65536K

Total Submissions: 51837 Accepted: 21619

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.

题目大意：定义字符串a*b等于”a+b”，给出一个字符串s问是否存在a^n=s,求出最大的n。

思路：kmp求最短循环节。如果i个字符组成一个周期串，那么“错位”部分（i-Next[i]）恰好是一个循环节。

#include<cstdio>
#include<cstring>
using namespace std;

const int maxn=1000005;
char s[maxn];
int Next[maxn];

void getNext()
{
    int n=strlen(s);
    Next[0]=Next[1]=0;
    for(int i=1;i<n;i++)
    {
        int j=Next[i];
        while(j&&s[i]!=s[j]) j=Next[j];
        if(s[i]==s[j]) Next[i+1]=j+1;
        else Next[i+1]=0;
    }
}

int main()
{
    while(~scanf("%s",s)&&s[0]!='.')
    {
        getNext();
        int n=strlen(s);
        if(Next[n]>0&&n%(n-Next[n])==0) printf("%d\n",n/(n-Next[n])); 
        else printf("1\n");
    }
    return 0;
}