poj2142 The Balance

The Balance

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 7437		Accepted: 3274

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

Source

Japan 2004

题目大意：给你两个标度的砝码，要求用最少数量的砝码称出物品的重.

分析：将题目转化为二元一次不定方程：ax+by=c,利用扩展欧几里得变形求解.如果x或y小于0，那么这个肯定就放在物品这一边.为了使得|x|+|y|最小，先让y取最小正整数解，看x为多少.再让x取最小正整数解，看y为多少，最后比较一下总和哪个大哪个小就好了.根据的是函数的单调性.

#include <cstdio>
#include <cmath>
#include <map>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long ll;

ll a, b, c, x3, y3, x4, y4, ansx, ansy;

ll exgcd(ll a, ll b, ll &x, ll &y)
{
    if (!b)
    {
        x = 1;
        y = 0;
        return a;
    }
    ll temp = exgcd(b, a%b, x, y), t = x;
    x = y;
    y = t - (a / b) * y;
    return temp;
}

int main()
{
    while (scanf("%lld%lld%lld", &a, &b, &c) && a + b + c)
    {
        ll x, y, d;
        d = exgcd(a, b, x, y);
        x = x * c / d;
        y = y * c / d;
        ll mod1 = b / d, mod2 = a / d;
        x3 = (x % mod1 + mod1) % mod1;
        y3 = (c - a*x3) / b;
        if (y3 < 0)
            y3 = -y3;
        y4 = (y % mod2 + mod2) % mod2;
        x4 = (c - b * y4) / a;
        if (x4 < 0)
            x4 = -x4;
        if (x3 + y3 < x4 + y4)
            ansx = x3, ansy = y3;
        else
            ansx = x4, ansy = y4;
        printf("%lld %lld\n", ansx, ansy);
        
    }

    return 0; 
}