题目

思路：要找到满足
x+y<=n
x⋅w+y⋅d=p,
->xw+yd+wd-wd=w*(x+d)+d*(y-w)
因为w>d，令x1=x+d,y1=y-w，x1+y1<x+y
只要x⋅w+y⋅d=p成立，w*(x+d)+d*(y-w)也成立，一旦y>w就可以化为y1=y%w使得x1+y1更小的情况，所以只要枚举y=[0,w-1]的所有情况即可

Code:

#include<iostream>
typedef long long ll;
using namespace std;
const int Max = 1e6 + 5;
int a[Max], b[Max];

int gcd(int a, int b)
{
	if (a == 0)return b;
	return gcd(b % a, a);
}

int main()
{
	ll n, p, w, d;
	cin >> n >> p >> w >> d;
	int f = 0;
	for (int i = 0;i <= 1e5;i++)
	{
		if (i * d > p)break;
		
		if ((p - i * d) % w == 0)
		{
			if (i + (p - i * d) / w <= n)
			{
					f = 1;cout << (p - i * d) / w << " " << i << " " << n - (p - i * d)/w - i;
				break;
			}
		}
	}
	if (f == 0)cout << -1 << endl;
}