CF 1244 C - The Football Season

C - The Football Season

先考虑求解

\[x\times w + y\times d=p \]

若存在一组解

\[\begin{cases} x_0\\ y_0 = kw + v & (0<v<w)\\ \end{cases} \]

则

\[x_0 \times w + y_0 \times d = p\\ \Rightarrow x_0 \times w + (kw+v)\times d = p\\ \Rightarrow x_0\times w + k\times w\times d + v\times d = p\\ \Rightarrow (x_0+k\times d)\times w + v\times d = p ~~~~~ \]

所以一定存在一组解

\[\begin{cases} x = x_0 + k\times d\\ y = v \end{cases} \]

并且由于\(w> d\)

有

\[x + y = x_0 + k\times d + v < x_0 + k\times w + v \]

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\n",x,v,n-x-v);
                return 0;
            }
        }
    }
    puts("-1");
    return 0;
}