poj 2406 Power Strings 哈希

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 61612		Accepted: 25463

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.

 

#include 
  
using namespace std;
typedef unsigned long long ll;
const int maxn = 1e6 + 100;
const int temp = 31;
string s;
ll f[maxn], Hash[maxn];
  
int main() {
    f[0] = 1;
    for (int i = 1; i < maxn; i++)
        f[i] = f[i - 1] * temp;
    while (cin >> s) {
        if (s[0] == '.') break;
        int l = s.length();
        Hash[l] = 0;
        for (int i = l - 1; i >= 0; i--) {
            Hash[i] = Hash[i + 1] * temp + (s[i] - 'a') + 1;
        }
        int ans = 0;
        for (int i = 1; i <= l; i++) {
            if (l % i != 0) continue;
            ll ha = Hash[0] - Hash[i] * f[i];
            int k = 0;
            for (k = i; k < l; k = k + i) {
                if (ha != Hash[k] - Hash[k + i] * f[i]) break;
                else ha = Hash[k] - Hash[k + i] * f[i];
            }
            if (k == l) {
                ans = l / i;
                break;
            }
        }
        cout << ans << endl;
    }
    return 0;
}