2142 The Balance

最新推荐文章于 2019-08-17 21:05:00 发布

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本文介绍了一种解决特定药品剂量测量问题的算法。通过使用两种不同重量的砝码，在确保最少砝码数量的前提下，实现精确测量指定重量的目标。文章详细阐述了输入输出格式及示例，并提供了完整的代码实现。

The Balance

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 1643		Accepted: 695

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

Source

Japan 2004

#include<stdio.h>
#include<iostream>
#include<math.h>
using namespace std;
int exgcd(int a,int b,int &x,int &y)
{
    int d,t;
    if(b==0)
    {
        x=1;
        y=0;
        return a;
    }
    d=exgcd(b,(a%b+b)%b,x,y);
    t=x;
    x=y;
    y=t-(a/b)*y;
    return d;
}
int main()
{
    int a,b,d,enda,endb;
    while(scanf("%d%d%d",&a,&b,&d)!=EOF)
    {
        if(a==0&&b==0&&d==0)    break;
        int f=0;
        if(a<b) swap(a,b),f=1;//转换，必须知道A,B哪个大，因为第一个要求就是要让砝码数量尽可能少
        int x,y;
        int dd=exgcd(a,b,x,y);
        x=x*(d/dd);
        y=y*(d/dd);
        int sa=a/dd,sb=b/dd;
        while(y<=0) x-=sb,y+=sa;//转换，为了使砝码数量最小
        int sum=1<<31-1,max=1<<31-1;
        while(y>-sa)
        {
            if(abs(x)+abs(y)<sum)
            {
                sum=abs(x)+abs(y);
                max=a*abs(x)+b*abs(y);//总质量最小
                enda=x;
                endb=y;
            }
            else if(abs(x)+abs(y)==sum)
            {
                 if(a*abs(x)+b*abs(y)<max)
                {
                    max=a*abs(x)+b*abs(y);
                    enda=x;
                    endb=y;
                }
            }
            y-=sa;
            x+=sb;
        }
       if(f) swap(enda,endb);
        printf("%d %d/n",abs(enda),abs(endb));
    }
    return 0;
}