Uva 10090 Marbles 扩展欧几里得 费用最小

/*
*   url: http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1031
*   stratege: 求n1*x+n2*y=n, x*c1+y*c2的值最小, 扩展欧几里得，分2种情况考虑,x的最小正整数解, y的最小正整数解
*   status：10090	Marbles	Accepted	C++	0.020	2012-08-30 08:10:38
*   Author: johnsondu
*/

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
using namespace std ;

#define LL long long 

void Euclid (LL a, LL b, LL &d, LL &x, LL &y)
{
    if (b == 0)
    {
        d = a ;
        x = 1 ;
        y = 0 ;
        return ;
    }
    Euclid (b, a%b, d, x, y) ;
    LL tmp = x ;
    x = y ;
    y = tmp - a/b*y ;
}

LL Gcd (LL x, LL y)
{
    return y ? Gcd (y, x%y) : x ;
}

int main ()
{
    LL n, n1, c1, n2, c2, d, x, y, x0, y0, mins, tx, ty, t;
    while (scanf ("%lld", &n), n)
    {
        scanf ("%lld%lld", &c1, &n1) ;
        scanf ("%lld%lld", &c2, &n2) ;
        
        t = Gcd (n1, n2) ;
        if (n%t)
        {
            printf ("failed\n") ;
            continue ;
        }
        
        n1 /= t, n2 /= t, n /= t ;
        Euclid (n1, n2, d, x, y) ;
        
        mins = -1 ;
        x *= n ;
        tx = x0 = ((x%n2)+n2) % n2 ;
        ty = y0 = (n-x0*n1)/n2 ;
        if (y0 >= 0)         //此时y0的为负数不可取
            mins = (x0*c1 + y0*c2) ;
        
        y *= n ;
        y0 = ((y%n1)+n1) % n1 ;
        x0 = (n-y0*n2)/n1 ;

        if ((mins > x0*c1 + y0*c2 || mins == -1) && x0 >= 0)  //x0为负数不可取
        {
            mins = (x0*c1 + y0*c2) ;
            tx = x0 ;
            ty = y0 ;
        }
        if (mins == -1)
            printf ("failed\n") ;
        else 
            printf ("%lld %lld\n", tx, ty) ;
    }
    return 0 ;
}