将 x 分解质因子后判断奇偶即可。
#include<bits/stdc++.h> #define int long long #define double long double #define x first #define y second using namespace std; typedef long long LL; typedef long long ll; typedef pair<int, int> PII; const int N = 3e5 + 10; const int M = 1e3 + 10; int mod = 1e9 + 7; int a[N]; void solve() { int n; cin >> n; int sum = 0; for (int i = 2; i * i <= n; i++) { while (n % i == 0) n /= i, sum++; } if (n > 1) sum++; if (sum % 2) cout << "kou\n"; else cout << "yukari\n"; } signed main() { ios::sync_with_stdio(false); cin.tie(0), cout.tie(0); int _; _ = 1; cin >> _; while (_--) { solve(); } }