UVA

Description

Problem E: Perfect Pth Powers

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 integerb, x = b^3. More generally, x is a perfect pthpower if, for some integer b, x = b^p. Given aninteger x you are to determine the largest p such that xis a perfect pth power.

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 therange of a (32-bit) int inC, C++, and Java. A line containing 0 follows the last test case.

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

Sample Input

17
1073741824
25
0

Output for Sample Input

1
30
2

题意：求x=b^p,的 最大的p

思路：首先将n分解质因数，然后找因数个数的gcd就是了，负数的话，一直除以2。直到答案是奇数的时候

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
typedef long long ll;
using namespace std;
const int maxn = 333333;

ll n;
int prime[maxn], vis[maxn], cnt = 0;

int gcd(int a, int b) {
	return b==0?a:gcd(b, a%b);
}

int solve() {
	ll tmp = n;
	if (n < 0)
		n = -n;
	int ans = 0;
	for (int i = 0; i < cnt && prime[i] <= n; i++) {
		int count = 0;
		while (n % prime[i] == 0) {
			count++;
			n /= prime[i];
		}
		ans = gcd(ans, count);
	}
	if (ans == 0)
		ans = 1;
	if (tmp < 0) {
		while (ans % 2 == 0)
			ans /= 2;
	}
	return ans;
}

int main() {
	for (int i = 2; i < maxn; i++) {
		if (vis[i])
			continue;
		prime[cnt++] = i;
		for (int j = i; j < maxn; j += i)
			vis[j] = 1;
	}
	while (scanf("%lld", &n) != EOF && n) {
		printf("%d
", solve());
	}
	return 0;
}