[Codeforces 1244C] The Football Season（擴展歐幾里得）

Codeforces 1244C

題意：給你n,p,w,d (w>d)
要你求一組x,y,z滿足上述倆個式子。

題解：因爲w>d,所以優先分配給x，x=p/w; 然後在分配給y，最後剩下的看它能不能整除gcd(w,d);
如果能，用擴展歐幾里德求解；然後加上去，最後 如果不滿足條件可以 對a和b的大小進行調整，
可以加上a*（m%d/gcd）個w和減去b*（m%d/gcd）個d 或者減去a*（m%d/gcd）個w和加上b*（m%d/gcd）個d。

#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
const ll mod=1e9+7; 

ll exgcd(ll a, ll b, ll &x, ll &y)  
{
    if(b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    ll d = exgcd(b, a % b, x, y);
    ll t = x;
    x = y;
    y = t - (a / b) * y;
    return d;
}
int main()
{
	ll n,p,w,d,x,y,z;
	ios::sync_with_stdio(false);
	cin>>n>>p>>w>>d;
	ll sum=p;
	ll gcd=__gcd(w,d);
	ll a=sum/w;
	sum%=w;
	ll b=sum/d;
	sum%=d;
	if(sum%gcd!=0)
		cout<<-1<<endl;
	else
	{
		//cout<<2<<endl;
		//ll cnt=sum/gcd;
		if(sum)
		{
				ll cnt=sum/gcd;
				exgcd(w,d,x,y);
				a+=x*cnt;
				b+=y*cnt;
		}
		while(a+b>n&&b>=0||a<0)
		{
			a+=d/gcd;
			b-=w/gcd;
		}
		while(b<0)
		{
			b+=w/gcd;
			a-=d/gcd;
		}
		
		if(a>=0&&b>=0&&a+b<=n&&a*w+b*d==p)
		cout<<a<<" "<<b<<" "<<n-a-b<<endl;
		else
		cout<<-1<<endl;
		//	cout<<a<<" "<<b<<" "<<n-a-b<<endl;
	}
	return 0;
}
//627936103814 4254617095171609 45205 1927
//627936103814 4254617095171609 45205 1927
//94118284813 15672 533817803329