题解 | #游游的数值距离#

找到一组 使 最接近与 即可，由于 增长的很快，我们枚举 ，可以 求出符合条件的 ，最后将答案取 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], fac[20];
int check(int x) {
    if (x == 2) return 0;
    return 1;
}

void solve() {
    int n;
    cin >> n;
    fac[0] = 1;
    int minn = 1e18, ansx, ansy;
    for (int i = 1; i <= 20; i++) fac[i] = fac[i - 1] * i;
    for (int i = 1; i <= 20; i++) {
        int x = i, y;
        int sum = 0;
        if (!check(x)) continue;
        if (x == 1) sum = n, y = 1;
        else {
            int k1 = n / (fac[x] - 1), k2 = (n + fac[x] - 2) / (fac[x] - 1);
            if (!check(k1)) sum = abs(n - k2 * (fac[x] - 1)), y = k2;
            else if (!check(k2)) sum = abs(n - k1 * (fac[x] - 1)), y = k1;
            else {
                if (abs(n - k1 * (fac[x] - 1)) < abs(n - k2 * (fac[x] - 1))) sum = abs(
                                n - k1 * (fac[x] - 1)), y = k1;
                else sum = abs(n - k2 * (fac[x] - 1)), y = k2;
            }
        }
        if (sum < minn) {
            minn = sum;
            ansx = x;
            ansy = y;
        }
    }
    cout << ansx << " " << ansy << "\n";
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int _;
    _ = 1;
    //cin>>_;
    while (_--) {
        solve();
    }
}