hdu 5610 Baby Ming and Weight lifting

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)

 

直接暴力枚举

AC代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<queue>
#include<set>
#include<bitset>
#include<map>
#include<vector>
#include<stdlib.h>
using namespace std;
#define ll long long
#define eps 1e-10
#define MOD 1000000007
#define N 1000000
#define inf 1e12
int a,b,c;
int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d%d",&a,&b,&c);
        if(c&1){
            printf("Impossible
");
            continue;
        }
        int half = c/2;
        
        int ans = 10000000;
        int ans1,ans2;
        for(int i=0;i<=1001;i++){
            for(int j=0;j<=1001;j++){
                int cnt = i*a + j*b;
                if(cnt==half){
                    if(ans>i+j){
                        ans=min(ans,i+j);
                        ans1=i,ans2=j;
                    }
                }
            }
        }
        if(ans==10000000){
            printf("Impossible
");
            continue;
        }
        printf("%d %d
",ans1*2,ans2*2);
    }
    return 0;
}