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.
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
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 $a+b$)
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; }