[POJ2406]Power Strings

题目链接：http://poj.org/problem?id=2406

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

 

 

给你一个字串，这个串是里面的某子串循环n次得到的，求这个n。

KMP求循环节，两个公式：如果n % (n - pre[n]) == 0说明存在循环节，循环次数为n / (n - pre[n])

 

 

 

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <queue>
#include <map>
#include <stack>
#include <list>
#include <vector>

using namespace std;

const int maxn = 66666666;
int na, nb;
char a[maxn];
char b[maxn];
int pre[maxn];

//b是模式串，a是目标串
void getpre(char *b, int *pre) {
    int j, k;
    pre[0] = -1;
    j = 0;
    k = -1;
    while(j < nb) {
        if(k == -1 || b[j] == b[k]) {
            j++;
            k++;
            pre[j] = k;
        }
        else {
            k = pre[k];
        }
    }
}

int kmp() {
    int ans = 0;
    int i = 0;
    int j = 0;
    getpre(b, pre);
    while(i < na) {
        if(j == -1 || a[i] == b[j]) {
            i++;
            j++;
        }
        else {
            j = pre[j];
        }
        if(j == nb) {
            ans++;
        }
    }
    return ans;
}

int main() {
    freopen("in", "r", stdin);
    while(~scanf("%s", b) && strcmp(".", b)) {
        nb = strlen(b);
        getpre(b, pre);
        int loop = nb - pre[nb];
        if(nb % loop == 0) {
            printf("%d
", nb / loop);
        }
        else {
            printf("1
");
        }
    }
}