28分求调

#include <iostream>
using namespace std;
int main()
{
    int n,x,y=36401,k;
    cin>>n;
    for (k=1;y%364!=0&&y>36400;k++)
        y=n-1092*k;
    x=y/364;
    k--;
    cout<<x<<endl;
    cout<<k<<endl;
    return 0;
}