POJ 1730 Perfect Pth Powers

题目描述

We say that x is a perfect square if, for some integer b, x = b 2. Similarly, x is a perfect cube if, for some integer b, x = b 3. More generally, x is a perfect pth power if, for some integer b, x = b p. Given an integer x you are to determine the largest p such that x is a perfect p th power.

Input

Each test case is given by a line of input containing x. The value of x will have magnitude at least 2 and be within the range of a (32-bit) int in C, C++, and Java. A line containing 0 follows the last test case.

Output

For each test case, output a line giving the largest integer p such that x is a perfect p th power.

Sample Input

17
1073741824
25
0

Sample Output

1
30
2

题目大意

给你一个绝对值大于等于2的整数n，问它能构成a^b=n中最大的b是多少？

解题报告

1：先套质因数分解，算出因子等信息。

2：我们知道一个大于2的正整数可以分解为n = (p1^a1)*(p2^a2)*(p3^a3).....其中p1, p2等为质因子。要使指数ax最大，只需要对所有指数求一遍gcd就行了。

3：负数先用正整数求出答案，由于负数的指数不可能为偶数。所有持续对ans除2，直到ans为奇数。

代码

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
typedef long long ll;
ll cnt[100], fac[100];
ll getFac(ll n){
    ll len = 0, s = sqrt(n);
    memset(cnt, 0, sizeof(cnt));
    for(ll i=2; i<=s; ++i){
        if(n%i == 0){
            fac[len] = i;
            while(n%i == 0)n /= i, cnt[len]++;
            ++len;
        }
    }
    if(n != 1){
        fac[len] = n, cnt[len]++, ++len;
    }
    return len;
}
int main(){
    ll n;
    while(~scanf("%lld", &n) && n){
        ll ans = 0, flag = 1;
        if(n < 0)flag = -1, n = -n;
        ll len = getFac(n);
        for(ll i=0; i<len; ++i){
            ans = __gcd(cnt[i], ans);
        }
        if(flag == -1){
            while(ans%2==0)ans >>= 1;
        }
        printf("%lld\n", ans);
    }
    return 0;
}