完全P次方数 Perfect P-th Powers

容易发现当 $n \leq 1$ 时，由于 $n$ 在 int 范围内，所以 $p < 31$。考虑从小到大枚举 $p$ 并二分 $\sqrt [p]{n}$，判断是否是整数即可。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <cassert>
#include <string>
using namespace std;

using ll = long long;
int t;
ll n;

ll qpow(ll x, ll y)
{
	ll res = 1, base = x;
	bool isbigger = 0;
	while (y)
	{
		if (y & 1LL)
		{
			res = res * base;
			if (res > n || isbigger) return -1;
		}
		base = base * base;
		if (base > n) isbigger = 1;
		y >>= 1LL;
	}
	return res;
}

ll qpow2(ll x, ll y)
{
	ll res = 1, base = x;
	bool isbigger = 0;
	while (y)
	{
		if (y & 1LL)
		{
			res = res * base;
			if (res < n || isbigger) return -1;
		}
		base = base * base;
		if (base < n) isbigger = 1;
		y >>= 1LL;
	}
	return res;
}

int main()
{
	while (scanf("%lld", &n) && n)
	{
		if (n >= 0)
		{
			for (ll i = 31; i >= 1; i--)
			{
				ll l = 1, r = n;
				while (l <= r)
				{
					ll mid = l + r >> 1;
					ll p = qpow(mid, i);
					if (p == -1)
					{
						r = mid - 1;
					}
					else
					{
						if (p == n)
						{
							printf("%lld\n", i);
							goto Ed;
						}
						l = mid + 1;
					}
				}
			}
		}
		else
		{
			for (ll i = 31; i >= 1; i -= 2)
			{
				ll l = n, r = -1;
				while (l <= r)
				{
					ll mid = l + r >> 1;
					n = -n;
					ll p = qpow(-mid, i);
					if (p == -1 || -p < -n)
					{
						n = -n;
						l = mid + 1;
					}
					else
					{
						n = -n;
						if (-p == n)
						{
							printf("%lld\n", i);
							goto Ed;
						}
						r = mid - 1;
					}
				}
			}
		}
	Ed:;
	}
	return 0;
}