poj 2406 Power Strings

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本文针对PowerStrings编程题，介绍了一种高效求解字符串最小周期的方法。通过计算最小周期串的长度，利用字符串长度进行除法运算得出结果。适用于字符串长度不超过1百万字符的情况。

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问题描述：
Power Strings
Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 42056 Accepted: 17526
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

题目大意：
给出一个字符串，求出该串有多少个最小周期串。

解题思路：
求出字符串的最小周期串的长度，然后用给定字符串的长度除以最小周期串的长度即可。

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

const int maxn = 1000005;
char a[maxn];

int main() {
    while(scanf("%s", &a) != EOF) {
        if(a[0] == '.')
            break;
        int l = strlen(a);
        for(int j, i = 1; i <= l; i ++) {
            if(l % i == 0) {
                for(j = i; j < l; j ++) {
                    if(a[j] != a[j % i])
                        break;
                }
                if(j == l) {
                    cout << l / i << endl;
                    break;
                }
            }
        }
    }
    return 0;
}