hdu-2142(扩展欧几里得解不定方程)

以你之姓@ 2022-06-10 10:14 121阅读 0赞

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.  
![2142_1.jpg][]

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

题目大意:给你俩个砝码，重量a和重量b,让你求得使用最少的砝码数量且砝码总重量也最小(即在|x|+|y|最小条件下，a\*|x|+b\*|y|也最小),称出重物c.

题目分析:很明显我们可以列出方程a\*x+b\*y=c这个方程，但是怎么求出满足题意得x和y了。我们知道了通解x,y完全可以依靠枚举来找最小的，嘿嘿。

暴力出奇迹，而且时间很快，远远少于5000m

![Center][]

代码如下:

#include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<cstdlib>
    #define ll long long
    using namespace std;
    
    ll myabs(ll a)//abs()这个函数好像在poj上不支持long long  所以自己写了一个
    {
        if (a<0) return -a;
        else return a;
    }
    ll exgcd(ll a,ll b,ll &x,ll &y)
    {
        if (b==0) {
            x=1;
            y=0;
            return a;
        }
        ll r=exgcd(b,a%b,x,y);
        ll t=x;
        x=y;
        y=t-a/b*y;
        return r;
    }
    void linear_equation(ll a,ll b,ll c,ll &x,ll &y)
    {
        ll d=exgcd(a,b,x,y);
        if (c%d)
            return ;
        ll k=c/d;
        x=x*k,y=y*k;
        ll t1=b/d,x1;
        ll t2=a/d,y1;
        ll Min=myabs(x)+myabs(y);
        ll ansx=x,ansy=y;
        for (ll i=-100000;i<=100000;i++) {//枚举x与y,根据判断条件来更新
            x1=x+t1*i;
            y1=y-t2*i;
            if (myabs(x1)+myabs(y1)<Min||(myabs(x1)+myabs(y1)==Min&&(myabs(x1)*a+myabs(y1)*b<myabs(ansx)*a+myabs(ansy)*b))) {
                ansx=x1;
                ansy=y1;
                Min=myabs(x1)+myabs(y1);
            }
        }
        printf("%lld %lld\n",myabs(ansx),myabs(ansy));
    }
    int main()
    {
        ll a,b,d,x,y;
        while (scanf("%lld%lld%lld",&a,&b,&d)!=EOF) {
            if (a==0&&b==0&&d==0) break;
            linear_equation(a,b,d,x,y);
        }
        return 0;
    }

[2142_1.jpg]: http://poj.org/images/2142_1.jpg
[Center]: /images/20220610/8213612f71f3427c828f2cfb0f9c38d7.png