HDU 5610 Baby Ming and Weight lifting 暴力

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

 

中文意思大概是给你一个可以放杠铃片的杠铃，相当于一个可以在两边放重物的杠杆。。。。

给你两个重物的质量（数量无限），问是否能使整个杠杆的重量为C。

一开始还以为要动规，发现数据范围只有1000，所以直接暴力枚举过了。

不过要让大数从大到小枚举，毕竟多数解时输出最小的那个。

代码如下：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<iostream>
using namespace std;
int n,a,b,c;
bool bo,bo1;
int main(){
    scanf("%d",&n);
    while(n--)
    {
        bo=false;
        bo1=false;
        scanf("%d %d %d",&a,&b,&c);
        if (c%2!=0)
        printf("Impossible\n");
        else
        {
            c=c/2;
        if (a>b)
        {
        swap(a,b);
        bo1=true;
        }
        for(int i=1000;i>=0;--i)
        {
            for(int j=1000;j>=0;--j)
            {
                  if (b*i+a*j==c&&bo1==true)
                  {
                  printf("%d %d\n",2*i,2*j);
                  bo=true;
                  break;
                  }else if(b*i+a*j==c){
                  
                  printf("%d %d\n",2*j,2*i);
                bo=true;
                break;
    }
            }
            if (bo==true)
            break;
        }
        if (bo!=true)
        printf("Impossible\n");
    }
    }
    return 0;
}