POJ 2406 Power Strings kmp

最新推荐文章于 2019-04-08 16:02:07 发布

蒟蒻的ACMer

最新推荐文章于 2019-04-08 16:02:07 发布

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本文介绍了一种用于解决字符串幂次分解问题的高效算法。该算法通过构建字符串的前缀函数来快速确定输入字符串是否可以表示为某个较短字符串的多次重复，并给出了完整的C++实现代码。

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Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 37817		Accepted: 15653

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

AC code：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#define maxn 1000005
using namespace std;
int next[maxn];
int main(){
    char s[maxn];
    while(scanf("%s",s)!=EOF&&strcmp(s,".")){
        int k=0,len=strlen(s),i,j;
        j=0,k=-1;
        next[0]=-1;
        while(j<len){
            if(k==-1||s[j]==s[k]){
			j++;k++;
			if(s[j]!=s[k])next[j]=k;
            else next[j]=next[k];
            }
            else k=next[k];
        }
        if(len%(len-next[len])==0)
            cout<<(len/(len-next[len]))<<'\12';
        else printf("1\n");
    }
    return 0;
}