题解 | #Old Bill#

写于2024.3.18

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;

int main()
{
    int i, j, k, n, x, y, z, hightest, lowest; // hightest, lowest分别表示最高位和最低位，即万位和个位
    cin >> n;                                  // 有n只火鸡
    cin >> x >> y >> z;
    int min = z * 10 + y * 100 + x * 1000; // 计算0xyz0的值
    // int maxhigh, maxlow;
    for (hightest = 9; hightest > 0; hightest--)
    {
        for (lowest = 9; lowest >= 0; lowest--)
        {
            if ((min + hightest * 10000 + lowest) % n == 0) // 符合条件的情况，可以输出
            {
                // cout << (min + hightest * 10000 + lowest) << endl;
                cout << hightest << " " << lowest << " " << (min + hightest * 10000 + lowest) / n << endl;
                return 0;
            }
        }
    }
    cout << "0" << endl;

    return 0;
}