CF 1244 C - The Football Season

Posted 2021-03-26 1625--h

C - The Football Season

先考虑求解
[ x imes w + y imes d=p ]
若存在一组解
[ egin{cases} x_0y_0 = kw + v & (0<v<w)\end{cases} ]
则
[ x_0 imes w + y_0 imes d = p\Rightarrow x_0 imes w + (kw+v) imes d = p\Rightarrow x_0 imes w + k imes w imes d + v imes d = p\Rightarrow (x_0+k imes d) imes w + v imes d = p ~~~~~ ]
所以一定存在一组解
[ egin{cases} x = x_0 + k imes dy = v end{cases} ]
并且由于(w> d)

有
[ x + y = x_0 + k imes d + v < x_0 + k imes w + v ]
前者比后者更有可能成为答案

所以直接枚举所有的v即可

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n,p,w,d;
int main(){
    cin >> n >> p >> w >> d;
    for(ll v=0;v<w;v++){
        if((p - v * d) % w == 0){
            ll x = (p - v * d) / w;
            if(x >= 0 && x + v <= n){
                printf("%lld %lld %lld
",x,v,n-x-v);
                return 0;
            }
        }
    }
    puts("-1");
    return 0;
}