2142 The Balance 拓展欧几里德，求|x|+|y|最小值

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该博客介绍了如何使用拓展欧几里得算法找到满足特定条件的非负整数对(x, y)，使得|x|+|y|达到最小值。内容涉及输入输出示例以及问题的解决方案。" 104544082,9224215,UART串行接口控制器设计详解,"['嵌入式硬件', '通信接口', '电路设计']

The Balance

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 1628		Accepted: 691

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.

You can measure dmg using x many amg weights and y many bmg weights.
The total number of weights (x + y) is the smallest among those pairs of nonnegative integers satisfying the previous condition.
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

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
int gcd(int a,int b,int &x,int &y)
{
    if(b==0)
    {
        x=1;y=0;
        return a;
    }
    int r=gcd(b,a%b,x,y);
    int t=x;
    x=y;
    y=t-a/b*y;
    return r;
}
int main()
{
    int a,b,val;
    while(cin>>a>>b>>val)
    {
        if(a==0&&b==0&&val==0) break;
        int flag=0;
        if(a<b) {swap(a,b);flag=1;}
        int x0,y0;
        int d=gcd(a,b,x0,y0);
        x0*=val/d;y0*=val/d;//此题中一定有解 所以不用考虑无解情况
        //此题中求abs(x)+abs(y)的最大值
        int sa=a/d,sb=b/d;
        int x=x0,y=y0;
        //x=x0+sb*t,y=y0-sa*t;
        int sum=(1<<31)-1,maxn=(1<<31)-1;
            while(y<=0)x-=sb,y+=sa;
            int ansx,ansy;
            while(y>-sa)
            {
                if(abs(x)+abs(y)<sum)
                {
                    sum=abs(x)+abs(y);
                    maxn=a*x+b*y;
                    ansx=x;ansy=y;
                }
                else if(abs(x)+abs(y)==sum)
                {
                    if(a*x+b*y<maxn)
                    {
                        maxn=a*x+b*y;
                        ansx=x;ansy=y;
                    }
                }
                y-=sa,x+=sb;
            }
            if(flag) swap(ansx,ansy);
            printf("%d %d/n",abs(ansx),abs(ansy));
    }
    return 0;
}