• POJ 2142 The Balance


    The Balance
    Time Limit: 5000MS   Memory Limit: 65536K
    Total Submissions: 6325   Accepted: 2787

    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
    #include <cstdio>
    #include <algorithm>
    int exgcd(int a,int b,int &x,int &y){
        if(b==0){
            x=1; y=0;
            return a;
        }
        int ans=exgcd(b,a%b,y,x);
        y-=a/b*x;
        return ans;
    }
    int main() {
        int a=1,b,c,x,y;
        bool Gavin;
        while (a || b || c){
            scanf("%d%d%d",&a,&b,&c);
            if (!a&&!b&&!c) break; //之前没有加这一句,输入最后的000结束时会继续运行一次,致命错误,被0除。
            Gavin=0;
            if(a<b){ 
            swap(a,b);Gavin=1;
            }
            int gcd=exgcd(a,b,x,y);
            a/=gcd;b/=gcd;c/=gcd;
            x*=c;y*=c;
            int t=y/a,x1,y1,x2,y2,ansx,ansy;
            while(y-a*t<0) t--;
            x1=abs(x+b*t);y1=abs(y-a*t);
            t++;
            x2=abs(x+b*t);y2=abs(y-a*t);
            if ((x1+y1<x2+y2)||(x1+y1==x2+y2)&&(x1*a+y1*b<x2*a+y2*b))
             ansx=x1,ansy=y1;
            else ansx=x2,ansy=y2;
            if(Gavin) printf("%d %d
    ",ansy,ansx);
            else printf("%d %d
    ",ansx,ansy);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/suishiguang/p/6366250.html
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