hdu 5616

Posted 2020-06-08

Baby Ming and Weight lifting

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 178    Accepted Submission(s): 77

Problem Description

Baby Ming is fond of weight lifting. He has a barbell pole(the weight of which can be ignored) and two different kinds of barbell disks(the weight of which are respectively a and b ), the amount of each one being infinite.

Baby Ming prepare to use this two kinds of barbell disks to make up a new one weighted C (the barbell must be balanced), he want to know how to do it.

 

 

Input

In the first line contains a single positive integer T , indicating number of test case.

For each test case:

There are three positive integer a,b , and C .

1≤T≤1000,0<a,b,C≤1000,a≠b

 

 

Output

For each test case, if the barbell weighted C can’t be made up, print Impossible.

Otherwise, print two numbers to indicating the numbers of a and b barbell disks are needed. (If there are more than one answer, print the answer with minimum a+b )

 

 

Sample Input

2

1 2 6

1 4 5

 

 

Sample Output

2 2

Impossible

 

 

Source

BestCoder Round #69 (div.2)

 

题意 ：t组数据 每组a b c

 两种杠盘 a，b

 目标质量 c

 

杠铃是成双出现的，然后只要枚举就可以了  注意优化

 

#include<iostream>
#include<cstdio>
#define LL __int64
using namespace std;
int t;
int a,b,c;
int flag;
int tt,ansa,ansb;
int main()
{
    while(scanf("%d",&tt)!=EOF)
    {
        for(int i=1; i<=tt; i++)
        {
            flag=0;
            scanf("%d%d%d",&a,&b,&c);
            if(c%2)
                printf("Impossible\n");
            else
            {
                if(a<b)
                {
                    t=a;
                    a=b;
                    b=t;
                    flag=1;
                }
                int gg=0;
                for(int j=0; j<=c/2/a; j++)
                    for(int k=0; k<=c/2/b; k++)
                    {
                        if(j*a+k*b==c/2)
                        {
                            ansa=j;
                            ansb=k;
                            gg=1;
                            break;
                        }
                    }
                if(gg==0)
                    printf("Impossible\n");
                else
                {
                    if(flag)
                        printf("%d %d\n",2*ansb,2*ansa);
                    else
                        printf("%d %d\n",2*ansa,2*ansb);
                }

            }


        }
    }

    return 0;
}