poj p2142——The Balance(扩展欧几里得算法)(复习)

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


题目大意：给出a和b，求ax+by=c的解且使|x|+|y|尽量小。若有多组解，则输出ax+by最小的x,y。
扩展欧几里得算法求出一组解，通过x=x0+kb',y=y0-ka'(a',b'为a和b除了gcd过后的值)枚举所有解。枚举范围：-30000~30000即可。若不想枚举，可以通过求f(x)得出当(x,y)在靠近或就在两坐标轴上时取最值(若判断斜率，则可以确定是x坐标还是y坐标)，然后直接比较即可。
祝自己NOIP顺利~

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
using namespace std;
int x,y,w,a,b;
int kzgcd(int a,int b,int &x,int &y)
{
    if(b==0)
    {
        x=1;y=0;
        return a;
    }
    int p=kzgcd(b,a%b,x,y);
    int t=x-a/b*y;
    x=y;y=t;return p;
}
int main()
{
    while(scanf("%d%d%d",&a,&b,&w)==3)
    {
        if(a==0&&b==0&&w==0)break;
        int g=kzgcd(a,b,x,y);
        int qwe=w/g;
        x*=qwe;y*=qwe;
        int k1=a/g,k2=b/g;
        int ansx=1e9+7,ansy=1e9+7,tot=1e9+7,weight=1e9+7;
        for(int i=-30000;i<=30000;i++)
        {
            int xx=x+i*k2,yy=y-i*k1;
            if(xx<0)xx=-xx;
            if(yy<0)yy=-yy;
            if(xx+yy<tot)
            {
                tot=xx+yy;
                ansx=xx;ansy=yy;
                weight=xx*a+yy*b;
            }
            if(xx+yy==tot&&(xx*a+yy*b<weight))
            {
                ansx=xx;ansy=yy;
                weight=xx*a+yy*b;
            }
        }
        printf("%d %d\n",ansx,ansy);
    }
    return 0;
}