【POJ 2142】The Balance

【题目】

传送门

Description

Ms. Iyo Kiffa-Australis has a balance and only two kinds of weights to measure a dose of medicine. For example, to measure 200mg of aspirin using 300mg weights and 700mg weights, she can put one 700mg weight on the side of the medicine and three 300mg weights on the opposite side (Figure 1). Although she could put four 300mg weights on the medicine side and two 700mg weights on the other (Figure 2), she would not choose this solution because it is less convenient to use more weights. 
You are asked to help her by calculating how many weights are required. 

Input

The input is a sequence of datasets. A dataset is a line containing three positive integers a, b, and d separated by a space. The following relations hold: a != b, a <= 10000, b <= 10000, and d <= 50000. You may assume that it is possible to measure d mg using a combination of a mg and b mg weights. In other words, you need not consider "no solution" cases. 
The end of the input is indicated by a line containing three zeros separated by a space. It is not a dataset.

Output

The output should be composed of lines, each corresponding to an input dataset (a, b, d). An output line should contain two nonnegative integers x and y separated by a space. They should satisfy the following three conditions. 
1、You can measure dmg using x many amg weights and y many bmg weights. 
2、The total number of weights (x + y) is the smallest among those pairs of nonnegative integers satisfying the previous condition. 
3、The total mass of weights (ax + by) is the smallest among those pairs of nonnegative integers satisfying the previous two conditions.
No extra characters (e.g. extra spaces) should appear in the output.

Sample Input

700 300 200
500 200 300
500 200 500
275 110 330
275 110 385
648 375 4002
3 1 10000
0 0 0

Sample Output

1 3
1 1
1 0
0 3
1 1
49 74
3333 1

 

【分析】

大致题意：现有一天平，以及质量为 a 和 b 的砝码，已知砝码数量不限且天平左右均可放砝码，现要求在天平上称出质量为 c 的物品。两种砝码可以分开放两边也可以放在同一边。求一种可行方案。要求：放置的砝码数量尽可能少；当砝码数量相同时，总质量尽可能小。

首先，答案最优的情况下两边不会放同一种砝码，那样的话它们会抵消掉

这样的话设质量为 a 的砝码放 x 个，质量为 b 的砝码放 y 个，那就有方程 ax + by = c

很明显就是扩欧了，我们可以先求出一组可行解，然后调整，使得 尽量小，在前者尽量小时让  尽量小。那么很容易想到最优情况要么 尽量小，要么 尽量小

调整的代码写得有点丑，大家自行领悟一下就行了

 

【代码】

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int gcd(int a,int b)
{
	int r=a%b;
	while(r!=0)
	{
		a=b;
		b=r;
		r=a%b;
	}
	return b;
}
void exgcd(int a,int b,int &x,int &y)
{
	if(!b)
	{
		x=1;
		y=0;
		return;
	}
	exgcd(b,a%b,y,x);
	y-=a/b*x;
}
void revise_x(int a,int b,int &x,int &y)
{
	while(x>0)  x-=b,y+=a;
	while(x<0)  x+=b,y-=a;
	if(abs(x-b)+abs(y+a)<abs(x)+abs(y))  x-=b,y+=a;
	if(abs(x+b)+abs(y-a)<abs(x)+abs(y))  x+=b,y-=a;
}
void revise_y(int a,int b,int &x,int &y)
{
	while(y>0)  x+=b,y-=a;
	while(y<0)  x-=b,y+=a;
	if(abs(x-b)+abs(y+a)<abs(x)+abs(y))  x-=b,y+=a;
	if(abs(x+b)+abs(y-a)<abs(x)+abs(y))  x+=b,y-=a;
}
int main()
{
	int a,b,c,x,y,p,q,r;
	while(~scanf("%d%d%d",&a,&b,&c))
	{
		if(!a&&!b&&!c)
		  break;
		r=gcd(a,b);
		a/=r,b/=r,c/=r;
		exgcd(a,b,x,y);
		x*=c,y*=c,p=x,q=y;
		revise_x(a,b,x,y);
		revise_y(a,b,p,q);
		if(abs(x)+abs(y)>abs(p)+abs(q))
		  x=p,y=q;
		printf("%d %d\n",abs(x),abs(y));
	}
	return 0;
}