题解 | #游游的数值距离#

枚举所有 x ，可能的 y 最多只有三个，x 也很少因为是阶乘

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 2e5 + 5;
int __t = 1, n, y, fac = 2, ans, ansx = 1, ansy = 1;
void ck(int x, int y) {
    if (y == 2 || y == 0)
        return;
    if (abs((fac - 1) * y - n) < ans) {
        ans = abs((fac - 1) * y - n);
        ansx = x;
        ansy = y;
    }
    return;
}
void solve() {
    cin >> n;
    ans = n;
    for (int x = 3; fac <= n; x++) {
        fac *= x;
        y = n / (fac - 1);
        ck(x, y - 1);
        ck(x, y);
        ck(x, y + 1);
    }
    cout << ansx << ' ' << ansy << endl;
    return;
}
int32_t main() {
#ifdef ONLINE_JUDGE
    ios::sync_with_stdio(false);
    cin.tie(0);
#endif
    // cin >> __t;
    while (__t--)
        solve();
    return 0;
}