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POJ2142(扩展欧几里得)

2024-08-12 10:02:38   220人阅读

The Balance
Time Limit: 5000MS Memory Limit: 65536K
Total Submissions: 5991 Accepted: 2605

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. 
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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 200500 200 300500 200 500275 110 330275 110 385648 375 40023 1 100000 0 0

Sample Output

1 31 11 00 31 149 743333 1
#include <cstdio>#include <algorithm>using namespace std;typedef long long LL;LL extgcd(LL a,LL b,LL &x,LL &y){    LL d=a;    if(b!=0)    {        d=extgcd(b,a%b,y,x);        y-=(a/b*x);    }    else    {        x=1;        y=0;    }    return d;}LL GCD(LL a,LL b){    if(b==0)    {        return a;    }    return GCD(b,a%b);}LL a,b,d;int main(){    while(scanf("%lld%lld%lld",&a,&b,&d)!=EOF&&(a+b+d)!=0)    {        LL x,y;        LL gcd=GCD(a,b);        a/=gcd;        b/=gcd;        d/=gcd;        extgcd(a,b,x,y);        x*=d;        y*=d;                LL x1=x;        x1=(x1%b+b)%b;//ax+by=1最小正整数解         LL y1=(d-a*x1)/b;        if(y1<0)y1=-y1;                LL y2=y;            y2=(y2%a+a)%a; //最小正整数解         LL x2=(d-b*y2)/a;        if(x2<0)x2=-x2;                if(x1+y1<x2+y2)        {            printf("%lld %lld\n",x1,y1);        }        else        {            printf("%lld %lld\n",x2,y2);        }    }        return 0;}

 

POJ2142(扩展欧几里得)


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