POJ 2142 The Balance

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题目链接：http://poj.org/problem?id=2142

题意：给两种砝码a,b，和待称量物品的重量d，用x个a和y个b来将物品称出，并使得x+y最小。

思路：a * x + b * y = d，扩展欧几里得。先让x为最小非负整数解，算出y；再让y为最小非负整数解，算出x，两种情况取最小的即可。

#include <cstdio>#include <cstring>#include <string>#include <cstdlib>#include <iostream>#include <algorithm>#include <stack>#include <map>#include <set>#include <vector>#include <sstream>#include <queue>#include <utility>using namespace std;#define rep(i,j,k) for (int i=j;i<=k;i++)#define Rrep(i,j,k) for (int i=j;i>=k;i--)#define Clean(x,y) memset(x,y,sizeof(x))#define LL long long#define ULL unsigned long long#define inf 0x7fffffff#define mod %100000007int a,b,d;int exgcd(int a,int b,int &x,int &y){    if ( b == 0 )    {        x = 1;        y = 0;        return a;    }    int r = exgcd( b , a % b , x , y );    int t = x;    x = y;    y = t - a/b * y;    return r;}void slove(int a,int b,int d){    int x,y;    int gcd = exgcd( a , b , x , y );    x*=d/gcd;    y*=d/gcd;    int addx = b/gcd;    int addy = a/gcd;    int m1,m2;    int ansx,ansy;    m1 = m2 = inf;    x = ( x % addx + addx )% addx;    y = (d - a*x)/b;    ansx = x;    ansy = abs(y);    m1 = x + abs(y);    m2 = a*x + b*abs(y);    y = ( y % addy + addy )% addy;    x = (d - b*y)/a;    if ( abs(x) + abs(y) < m1 )        {            m1 = abs(x) + abs(y);            ansx = abs(x);            ansy = abs(y);        }        else if ( abs(x) + abs(y) == m1 && a*abs(x) + b*abs(y) < m2 )        {            ansx = abs(x);            ansy = abs(y);        }    printf("%d %d\n",ansx,ansy);}int main(){    while( ~scanf("%d%d%d",&a,&b,&d) )    {        if ( a + b + d == 0 ) break;        slove(a,b,d);    }    return 0;}

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