给定n个整数和一个k,求有多少个数对(i,j)满足

\[a_i \times a_j = x^k \]

我们不妨对每个a进行质因数分解，发现两个数字相乘是某个数的k次方，就是两个数质因数分解后，对应质数的次幂和为k的倍数，这样我们可以hash统计来解决(用map)

#include<bits/stdc++.h>
#define rep(i,j,k) for(int i(j);i<=k;++i)
#define drp(i,j,k) for(int i(j);i>=k;--i)
#define repg(x) for(int i(G.head[x]);i;i=G.next[i])
#define bug cout<<"~~~~~~~~~~~~~"<<'\n';
using std::cin;
using std::cout;
typedef long long lxl;
template<typename T>
inline T  max( T a, T b) {
	return a > b ? a : b;
}
template<typename T>
inline T  min( T a, T b) {
	return a < b ? a : b;
}

inline char gt() {
	static char buf[1 << 21], *p1 = buf, *p2 = buf;
	return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;
}
template <typename T>
inline void  read(T &x) {
	char ch = gt();
	x = 0;
	int w(0);
	while(!(ch >= '0' && ch <= '9'))w |= ch == '-', ch = gt();
	while(ch >= '0' && ch <= '9')x = x * 10 + (ch & 15), ch = gt();
	w ? x = ~(x - 1) : x;
}
template <typename T>
inline void out(T x, char cc) {
	if(x < 0) x = -x, putchar('-');
	char ch[20];
	int num(0);
	while(x || !num) ch[++num] = x % 10 + '0', x /= 10;
	while(num) putchar(ch[num--]);
	putchar(cc);
}

const int N = 1e5 + 79;
int n, a, k;

std::vector<std::pair<int, int> > x, y;
std::map<std::vector<std::pair<int, int> >, int> mp;
lxl ans;

int prime[N], tot, v[N];

inline void EulerSieve(int MX) {
	rep(i, 2, MX) {
		if(!v[i]) {
			v[i] = i;
			prime[++tot] = i;
		}
		rep(j, 1, tot) {
			if(prime[j] > v[i] || 1ll * i * prime[j] > MX) break;
			v[i * prime[j]] = prime[j];
		}
	}
}

int main() {
	EulerSieve(100000);
	read(n);
	read(k);
	while(n--)	{
		read(a);
		x.clear();
		y.clear();
		rep(i, 1, tot) {
			if(prime[i]*prime[i] > a)	 break;
			int cnt(0);
			while(a % prime[i] == 0) {
				++cnt;
				a /= prime[i];
				cnt %= k;
			}

			if(!cnt) continue;
			x.push_back({prime[i], cnt});
			y.push_back({prime[i], k - cnt});
		}

		if(a != 1) { //±¾ÉíÊÇÒ»¸öÖÊÊý
			x.push_back({a, 1});
			y.push_back({a, k - 1});
		}
		ans += mp[y];
		++mp[x];
	}
	out(ans, '\n');
	return 0;
}