The Balance Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 3341   Accepted: 1471 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
题型：数论（求解模线性方程）
题意：有重为a、b两种砝码和重d的药品，现在用着两种砝码称出药品，问需要x个a砝码和y个b砝码，使x+y最小。
分析： 先求a * x ≡ d ( mod b )，得到x后根据x求得y，y = ( d - a * x ) / b； 相似地，再求解b * y ≡ d ( mod a )得到x和y 然后去x+y小的哪一组。 注意：若x或y是负的，要将其变正。
代码： #include #include #include #include #define LL long long using namespace std; int extgcd(int a,int b,int & x,int & y) { if(b==0) { x=1; y=0; return a; } int d=extgcd(b,a%b,x,y); int t=x; x=y; y=t-a/b*y; return d; } int modeq(int a,int b,int n) { int x,y,d,b1; d=extgcd(a,n,x,y); if(b%d!=0) return -1;//无解 b1=(n/d)<0 ? -n/d : n/d; x*=b/d; return (x%b1+b1)%b1; } int main() { int a,b,d; while(1) { scanf("%d%d%d",&a,&b,&d); if(a==0&&b==0&&d==0) break; int x1,y1,x2,y2; x1=modeq(a,d,b); y1=(d-a*x1)/b; if(x1<0) x1=-x1; if(y1<0) y1=-y1; y2=modeq(b,d,a); x2=(d-b*y2)/a; if(y2<0) y2=-y2; if(x2<0) x2=-x2; if(x1+y1