POJ 2406 Power Strings（KMP）

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.

 

题目大意：给一个字符串，问这个字符串是否能由另一个字符串重复R次得到，求R的最大值。

思路：对于一个字符串，如abcd abcd abcd，由长度为4的字符串abcd重复3次得到，那么必然有原字符串的前八位等于后八位。

也就是说，对于某个字符串S，长度为N，由长度为n的字符串s重复R次得到，当R≥2时必然有S[1..N-n]=S[n+1..N]。

那么对于KMP算法来说，就有fail[N]=N-n。此时n肯定已经是最小的了。

然后只需要判断N是否n的倍数，是则输出N/n即可。否则输出1。

 

代码（157MS）：

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

const int MAXN = 1000010;

void getFail(char *P, int m, int f[]) {
    f[0] = f[1] = 0;
    for(int i = 1; i < m; ++i) {
        int j = f[i];
        while(j && P[i] != P[j]) j = f[j];
        f[i + 1] = (P[i] == P[j] ? j + 1 : 0);
    }
}

char s[MAXN];
int fail[MAXN], n;

int main() {
    while(scanf("%s", s) != EOF) {
        if(*s == '.') break;
        n = strlen(s);
        getFail(s, n, fail);
        if(n % (n - fail[n]) == 0) printf("%d
", n / (n - fail[n]));
        else puts("1");
    }
}