拓展欧几里得——egcd poj2142
The Balance
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
第一次写拓展欧几里德的题，花了很久理解这个算法，搞清楚了这些egcd返回值就是gcd( a , b );函数里面算出来的x，y是随机传参的。算完了得到的x,y再*d/ gcd 才能得到初始的x,y.
这个题题意：求出满足a*x+b*y=d 的最小的 | x | + | y |;
先把x0,y0,算出来;
由x,y 的一般式
x=x0+b/d*t
y=y0-a/d*t 可以先假设 a>b,不满足就swap（a，b）
然后得到这么个B玩儿 |x|+|y|==|x0+b/d*t|+|y0-a/d*t|；
然后xjb猜当y0=a/d*t的时候 |x|+|y|可能最小；
最后在这个t的周围试几个数就行了……
#include
#include
#include
#include
using namespace std;
typedef long long ll;
const int INF=0x3f3f3f3f;
int a,b,d;
int egcd(int a,int b,int &x,int &y)
{
if(b==0)
{
x=1;
y=0;
return a;
}
int gcd=egcd(b,a%b,y,x);
y-=a/b*x;
return gcd;
}
int main()
{
while(scanf("%d%d%d",&a,&b,&d)!=EOF,a||b||d)
{
int x,y;
int flag=0;
if(ahh)
{
ans=hh;
minx=fabs(x+(b/gcd)*i);
miny=fabs(y-(a/gcd)*i);
//cout<