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自然万物都趋向从有序变得无序

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The Balance POJ - 2142

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题目链接:点我

题目

    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.

这里写图片描述

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

题意:

给你三个重量为a, b, c的物品,前两个物品可以使用无限个,但第三个物品只能而且必须使用一个,把他们放在天平上,求前两个物品使用的个数x, y,并且使得 x+y 的值最小.

思路:

即求ax + by = c的满足等式的使得x + y 的值最小的整数解,那么我们可以用扩展欧几里得算法,这里我们把方程两边同时除了gcd(a, b)即可满足扩展欧几里得算法的条件,然后把最后的答案乘以c / gcd(a,b)即可.

代码:

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#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<iostream>
using namespace std;

typedef long long LL;

int gcd(int a,int b){
    return b ? gcd(b, a%b) :a;
}

int exgcd(int a,int b,int &x,int &y){
    if(!b){
        x=1;
        y=0;
        return a;
    }else {
        int d=exgcd(b,a%b,y,x);
        y-=x*(a/b);
        return d;
    }
}

int main(){
    int a,b,c;
    while(scanf("%d %d %d", &a, &b, &c) != EOF &&(a+b+c)){
        int x,y;
        int g=gcd(a,b);
        a/=g;
        b/=g;
        c/=g;
         exgcd(a,b,x,y);
        int vx = x *c;
        vx= (vx%b+b)%b;
        int vy=(c-a*vx)/b;
        if(vy<0) vy=-vy;
        y=y*c;
        y=(y%a+a)%a;
        x=(c-b*y)/a;
        if(x<0) x=-x;
        if(x+y>vx+vy)
            x=vx,y=vy;
        printf("%d %d\n",x,y);
    }
    return 0;
}