【POJ 2406】Power Strings-优快云博客

本文介绍了一种解决字符串幂次分解问题的算法，利用KMP算法优化处理大规模输入数据，实现快速查找最大幂次。

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 64676		Accepted: 26679

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

题解：KMP……照着书打的qwq

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
using namespace std;
#define maxn 2000100
char s[maxn];
int nexts[maxn];
int n,m;
void getnexts()
{
    int j=0;
    int k=-1;
    nexts[0]=-1;
    while(j<m)
    {
        if(k==-1||s[j]==s[k])
        {
            j++;k++;
            nexts[j]=k;
        }
        else k=nexts[k];
    }
}
int main()
{
    while(~scanf("%s",s))
    {
        if(s[0]=='.')break;
        m=strlen(s);
        getnexts();
        int n=m-nexts[m];
        if(n!=m&&m%n==0)
            printf("%d\n",m/n);
        else
            printf("1\n");
    }
    return 0;
}