题解 | #游游的数值距离#

#include <iostream>
#define int long long
using namespace std;

signed main() {
    int n, a, b;
    cin >> n;
    if(n <= 2)cout << "1 1";
    else if(n == 4 || n == 3)cout << "3 1";
    else{
        int first = 2, minn = 999999999;
        for(int i = 3; i <= 13; i++){
            first *= i;
            int now = first - 1;
            int bb = n / now;
            //cout << i << ' ' << bb <<" " << n - now * bb << " " << now * (bb + 1) - n  << " ";
            if((n - now * bb) < minn && bb != 2 && bb != 0){
                a = i;
                b = bb;
                minn = n - now * bb;
            }
            if((now * (bb + 1) - n) < minn && bb != 1 && bb != 0){
                a = i;
                b = bb + 1;
                minn = now * (bb + 1) - n;
            }
            //cout << i << ' ' << minn << endl;
        }
        cout << a << ' ' << b;
    }
}
// 64 位输出请用 printf("%lld")

直接模拟做, 因为12!就比1e9大了,

对于每一个!考虑两种情况的b:第一个是不足n, 第二个是超过n,

注意第二种情况的b!=2判断不要写错了

(在这里想了很久的人)