poj 2406 Power Strings（KMP入门，next函数理解）

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 37685		Accepted: 15590

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

基础题目，只要是理解next函数。

#include<iostream>
#include<vector>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <math.h>
#include<algorithm>
#define ll long long
#define eps 1e-8
using namespace std;

int nexts[1050];
char b[1050];

void pre_nexts(int m)
{
    memset(nexts,0,sizeof(nexts));
    int j = 0,k = -1;
    nexts[0] = -1;
    while(j < m)
    {
        if(k == -1 || b[j] == b[k])  nexts[++j] = ++k;
        else k = nexts[k];
    }
}
int main(void)
{
    while(scanf("%s",b),b[0] != '.')
    {
        int n = (int)strlen(b);
        pre_nexts(n);
        if(n % (n-nexts[n]) == 0 && n/(n - nexts[n]) > 1) printf("%d\n",n/(n-nexts[n]));
        //根据next函数的最后一个匹配数，算出循环节点～
        else printf("1\n");
    }
    return 0;
}