题目链接:https://ac.nowcoder.com/acm/contest/4853/D
图片说明
思路:
图片说明

#include <bits/stdc++.h>
using namespace std;
#define ll long long

ll extend_gcd(ll a,ll b,ll&x,ll&y) {
    if(!b) {
        x = 1;
        y = 0;
        return a;
    }
    ll xt = 0, yt = 0;
    ll d = extend_gcd(b, a % b, xt, yt);
    x = yt;
    y = xt - yt * (a/b);
    return d;//gcd
}

ll minx(ll a,ll b,ll c)//x=(x%b+b)%b;(求出的就是最小正整数解)
{
    ll x,y;
    ll gcd=extend_gcd(a,b,x,y);
    if(c%gcd!=0) return -1;
    x*=c/gcd;//转化为a*x+b*y=c的解
    b/=gcd;//约去c后原来b就变为了b/gcd;
    if(b<0) b=-b;//如果b为负数就去绝对值
    ll ans=x%b;
    if(ans<=0) ans+=b;//求最小正整数解
    return ans;//返回的就是最小正整数解
}

int main(){
    ll a, b, c, k;
    ll x, y, z, w;
    cin>>a>>b>>c>>k;
    ll d=__gcd(b, c);
    x=minx(a, d, k);
    w=(k-x*a)/d;
    b/=d; c/=d;
    y=minx(b, c, w);
    z=(w-y*b)/c;
    cout<<x<<" "<<y<<" "<<z<<endl;

    return 0;
}