hdoj 5610 Baby Ming and Weight lifting 【暴力】【水题】

最新推荐文章于 2016-04-26 20:51:57 发布

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最新推荐文章于 2016-04-26 20:51:57 发布

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本题涉及BabyMing使用两种不同重量的杠铃盘实现特定总重量的平衡杠铃的问题。需要通过算法找出最少数量的杠铃盘组合以达到目标重量。

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Baby Ming and Weight lifting

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 727 Accepted Submission(s): 296

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

Sample Output

  
  
   
   2 2
Impossible

Source

BestCoder Round #69 (div.2)

由于杠铃要两边平衡，所以C是奇数的时候直接impossible~

AC代码：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#define mem(a, b) memset(a, (b), sizeof(a))
#define Wi(a) while(a--)
#define Si(a) scanf("%d", &a)
#define Pi(a) printf("%d\n", (a))
#define INF 0x3f3f3f3f
#include<algorithm>
using namespace std;
int main()
{
	int T; Si(T);
	Wi(T)
	{
		int a, b, c;
		scanf("%d%d%d", &a, &b, &c);
		if(c%2){
			printf("Impossible\n");
			continue;
		}
		else
		{
			int i, j;
			int ans = INF;
			int x, y;
			bool flag = 0;
			for(i = 0; i <= c/a; i++)
			{
				for(j = 0; j <= (c-a*i)/b; j++)
				{
					if(i*a+j*b == c/2)
					{
						flag = 1;
						if(ans > i+j )
						{
							ans = i+j;
							x = i, y = j;
						}	
					}
				}
			}
			if(!flag)  printf("Impossible\n");
			else	printf("%d %d\n", 2*x, 2*y);
		}
	}
	return 0;
}