The Balance（数论）

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

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

解题思路

拓展欧几里得求最小正解，比较难以理解的是求出一组解后，如何求最小正解的过程，此题再次再次求教了userluoxuan，在此鸣谢

AC代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
using namespace std;
const int INF = 999999999;
int gcd(int a, int b)
{
    return b ? gcd(b, a % b) : a;
}
int extend_gcd (int a, int b, int &x, int &y)
{
    if(b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    int r = extend_gcd(b, a % b, x, y);
    int temp = x;
    x = y;
    y = temp - a / b * y;
    return r;
}
int main()
{
    int a, b, n, d, x, y, min, min_x, min_y;
    while(scanf("%d%d%d", &a, &b, &n) && ( a || b || n) )
    {
        bool flag = false;
        min = INF;
        if(a < b)
        {
            swap(a, b);
            flag = true;                           //原方程等式为a * x + b * y = n 

        }
        d = gcd(a, b);                             //求a，b的最大公约数，a，b化成互质的
        a /= d;
        b /= d;
        extend_gcd(a, b, x, y);                    //求出a * x + b * y = n / gcd(a , b)  的一组解         
        x *= n / d;                                //化成原方程的解
        y *= n / d;
        int t = y / a;                             //由通解公式有 y(n) = y - a * t ，令y(n) = 0，有 t = y / a，此t为0点左右的
        for(int i = t - 1; i <= t + 1; i++)        //查找t左右两解，寻找最小正解
        {
            int x_1 = x + b * i;
            int y_1 = y - a * i;
            if(abs(x_1) + abs(y_1) < min)
            {
                min = abs(x_1) + abs(y_1);
                min_x = abs(x_1);
                min_y = abs(y_1);
            }
        }
        if(flag)
            printf("%d %d\n", min_y, min_x);
        else
            printf("%d %d\n", min_x, min_y);
    }
    return 0;
}