poj 2142 The Balance

 

扩展欧几里得的运用：

View Code

#include<iostream>

#include<cstdio>

#include<cstdlib>

#include<algorithm>

#include<cmath>

#include<queue>

#include<set>

#include<map>

#include<cstring>

#include<vector>

#define LL long long

using namespace std;

LL Ex_Gcd( LL a , LL b , LL &x , LL &y )

{

   if( b ==0 )

   {

       x = 1 ; y = 0;

       return a;        

   }

   LL mod = Ex_Gcd( b , a%b , x , y );

   LL temp = x;

   x = y;

   y = temp - ( a / b ) * y;

   return mod;

}

int main(  )

{

    LL A,B,C,x,y,X,Y;

    while( scanf( "%I64d %I64d %I64d",&A,&B,&C ),A||B||C )

    {

           LL mod = Ex_Gcd( A , B , x ,y );

           A /= mod;

           B /= mod;

           C /= mod;

          //当x是最小正整数解 

           X = ( (x*C)%B + B )%B;

           Y = ( C - X*A )/B;

           if( Y < 0 ) Y = -Y;

           //当y是最小正整数解  

           y = ( (y*C)%A + A )%A;

           x = ( C - y*B )/A;

           if( x < 0 ) x = -x;

           //判断和最小  

           if( x + y < X + Y )

           {

               X = x ; Y = y;

           }

           printf( "%I64d %I64d\n",X,Y);    

    }

    //system( "pause" );

    return 0;

}