HDU 5610 Baby Ming and Weight lifting

Baby Ming and Weight lifting

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1365    Accepted Submission(s): 500

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

 

题目大意：2种类型的杠铃片（重量分别为a和b），每种杠铃片都有无限个。用这2种杠铃片组成重量为C的杠铃（杠铃必须平衡）。求a, b的数量(若如果有多种答案，输出a+b最小的方案)

思路：大神的方法太屌，982MS过

 

 

#include <stdio.h>
int main()
{
        int i, j, t, a, b, c, flag;
        scanf("%d", &t);
        while(t--)
        {
        scanf("%d%d%d", &a, &b, &c);
        if(c&1)
        {
        printf("Impossible
");
        continue;
        }
        for(c /= 2,flag=0,i=0;!flag&&i<=2000;i++)//a+b的个数
                for(j=0;!flag&&j<=i;j++)//b的个数
                        if((i-j)*a+j*b==c&&(flag=1))
                        printf("%d %d
", (i-j)<<1, j<<1);
        if(!flag) printf("Impossible
");
        }
        return 0;
}