POJ2406 Power Strings —— KMP or 后缀数组 最小循环节

题目链接：https://vjudge.net/problem/POJ-2406

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 52631		Accepted: 21921

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

KMP：

求字符串的最小循环节。

1) 如果字符串长度能被“初步的最小循环节”整除，那么这个就是他的最小循环节。

2) 如果字符串长度不能被“初步的最小循环节”整除，那么这个只是它通过补全后的最小循环节。而它实际的最小循环节为自己。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <sstream>
#include <algorithm>
using namespace std;
typedef long long LL;
const double eps = 1e-6;
const int INF = 2e9;
const LL LNF = 9e18;
const int MOD = 1e9+7;
const int MAXN = 2e6+10;

char x[MAXN];
int Next[MAXN];

void get_next(char x[], int m)
{
    int i, j;
    j = Next[0] = -1;
    i = 0;
    while(i<m)
    {
        while(j!=-1 && x[i]!=x[j]) j = Next[j];
        Next[++i] = ++j;
    }
}

int main()
{
    while(scanf("%s", x) && strcmp(x, "."))
    {
        int len = strlen(x);
        get_next(x, len);
        int r = len-Next[len];  //求出最小循环节
        if(len%r)       //如果长度/最小循环节除不尽，则对于此字符串来说，实际的最小循环节是自己
            printf("1
");
        else
            printf("%d
", len/r);
    }
}

后缀数组：