題目鏈接:https://codeforces.com/problemset/problem/1247/D
題意:給你一個序列a,問有多少對i,j滿足存在x使得 ai * aj = x ^ k , k是給定的。
思路:兩個數滿足條件就是要相同因數的個數要是k的倍數,
把每個數質因數分解,然後算出來補數,例如還需要2個2,就是4
然後我們就在map裏找這個補數,然後再將當前數的貢獻存入map
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define ls rt << 1
#define rs rt << 1|1
#define mid ((l + r) >> 1)
#define lson l, mid, ls
#define rson mid + 1, r, rs
const int maxn = 1e5 + 10;
int aa[maxn], prime[maxn], a[maxn];
map<ll, int> mp;
int tot = 0;
void init()
{
for(int i = 2; i < maxn; ++i)
{
if(!aa[i])
{
prime[++tot] = i;
for(int j = i * 2; j < maxn; j += i)
aa[j] = 1;
}
}
}
int main()
{
init();
int n, k;
scanf("%d%d", &n, &k);
for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);
ll ans = 0;
for(int i = 1; i <= n; ++i)
{
queue<pair<int, int> > q;
while(!q.empty()) q.pop();
int x = a[i];
if(x == 1) q.push({1, 1});
for(int j = 1; j <= tot && prime[j] <= x; ++j) //質因數分解
{
int cnt = 0;
while(x % prime[j] == 0)
{
x /= prime[j];
++cnt;
}
cnt = cnt % k;
if(cnt) q.push({prime[j], cnt});
}
bool flag = true;
ll res = 1, res1 = 1;
while(!q.empty())
{
pair<int, int> tmp = q.front(); q.pop();
for(int j = 1; j <= k - tmp.second; ++j) //算補數
res = res * tmp.first;
for(int j = 1; j <= tmp.second; ++j) //算貢獻
res1 = res1 * tmp.first;
}
if(res > 100000) flag = false; //可以優化一下,超出值的範圍一定不行
if(flag) ans += mp[res];
mp[res1]++;
}
printf("%lld\n", ans);
return 0;
}