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&){
             printf("Impossible\n");
             continue;
         }
         int half = c/;

         int ans = ;
         int ans1,ans2;
         for(int i=;i<=;i++){
             for(int j=;j<=;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==){
             printf("Impossible\n");
             continue;
         }
         printf("%d %d\n",ans1*,ans2*);
     }
     return ;
 }