The Balance
Time Limit: 5000MS Memory Limit: 65536K
Total Submissions: 930 Accepted: 384

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

        怎么評價這道題呢，會做了，覺得這題挺水的，但是自己做了一下午，先是TLE后是WA，郁悶的不行，思維啊思維，做競賽考思維啊！
題目大意：
給定a,b兩種藥品的重量，用他們的組合來測量第三種重量為d的藥品，標(biāo)準(zhǔn)：
1.假設(shè)取x個重為a的藥品，取y個重為b的藥品；
2.在滿足以上情況下，使得x+y最??；
3.在滿足以上情況下，使得ax + by最小。
求x和y的值。（x和y均為正整數(shù)）

思路：
有三種情況：
ax+by=d;(a,b均在左盤)
ax-by=d;（a在左盤，b和d在右盤）
-ax+by=d;（b在左盤，a和d在右盤）
枚舉a，求滿足條件的解。

代碼：
#include<iostream>
#include<cmath>
using namespace std;
int main()
{
    int i,a,b,d,x,y,xx,yy;
    int sum1,sum2;
    while(cin>>a>>b>>d)
    {
        if(a==0 && b==0 && d==0)break;
        int min1=9000000,min2=9000000;
        int t=0;
        xx=0;yy=0;
        for(i=0; ; i++){
            if((d-a*i)%b==0)//ax+by=d 和 ax-by=d 兩種情況，用絕對值表示，所以合寫成一個if 
            {
                yy=abs((d-a*i)/b);
                sum1=i+yy;
                sum2=a*i+b*yy;
                if(sum1<min1)
                    min1=sum1,min2=sum2,x=i,y=yy;
                else if(sum1==min1 && sum2<min2)
                    min1=sum1,min2=sum2,x=i,y=yy;
                if(i>min1 || i*a>min2)//如果x的值比x+y的值都大或者a*x比a*x+b*y 的值都大了就跳出 
                    break;
            }
            int l=i*a+d;
            if(l>0 && l%b==0)//-ax+by=d的情況 
            {
                yy=l/b;    
                sum1=i+yy;
                sum2=a*i+b*yy;
                if(sum1<min1)
                    min1=sum1,min2=sum2,x=i,y=yy;
                else if(sum1==min1 && sum2<min2)
                    min1=sum1,min2=sum2,x=i,y=yy;
                if(i>min1 || i*a>min2)//如果x的值比x+y的值都大或者a*x比a*x+b*y 的值都大了就跳出
                    break;
                
            }
        }

        cout<<x<<" "<<y<<endl;

    }

return 0;
}