The Balance

The Balance

Meaning of the questions: a two weights and optimal solution an item (items may be about balance) seeking to make the balance equilibrium given. 
Interpretations: ax EQUATIONS + by = c, for x, y are taken positive and negative still on. Conversion for the sake of | x | + | y | optimal solution x = x0 + k * b / gy = y0-k * a / g; 
for | x0 + k * b / g | + | y0-k * a / G | optimal solution must be in x0 + k * b / g = 0 and y0-k * a / g = 0, between the two k, traverse just fine. (Piecewise)

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

void exgcd(ll a,ll b,ll &d,ll &x,ll &y)
{
    if (!b)
    {
        d=a;
        x=1;
        y=0;
        return;
    }
    exgcd(b,a%b,d,y,x);
    y-=a/b*x;
}

int main()
{
    ll a,b,c,d,x,y,ansx,ansy,xx,yy,zz,ans,t;
    while (scanf("%lld%lld%lld",&a,&b,&c))
    {
        if (a==0&&b==0&&c==0)
        {
            break;
        }
        exgcd(a,b,d,x,y);
        x=x*c/d;
        y=y*c/d;
        = 0x3f3f3f3f years;
        t=y/(a/d);
        zz=-x/(b/d);
        ansx=abs(x);
        ansy=abs(y);
        ans=ansx+ansy;
        for (ll i=min(zz,t)-1;i<=max(zz,t)+1;i++){
            yy=y-a/d*i;
            //printf("%lld\n",i);
            if (yy<0) yy=-yy;
            xx=x+b/d*i;
            if (xx<0) xx=-xx;
            if (ans>xx+yy){
                ans=xx+yy;
                ansx=xx;
                ansy=yy;
            }else{
                if (ans==xx+yy){
                    if (a*ansx+b*ansy>a*abs(xx)+b*abs(yy)){
                        ansx=xx;
                        ansy=yy;
                    }
                }
            }
        }
        printf("%lld %lld\n",ansx,ansy);
    }
    return 0;
}