POJ1017 箱子装填

Packets

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 38381		Accepted: 12817

Description

A factory produces products packed in square packets of the same height h and of the sizes 1*1, 2*2, 3*3, 4*4, 5*5, 6*6. These products are always delivered to customers in the square parcels of the same height h as the products have and of the size 6*6. Because of the expenses it is the interest of the factory as well as of the customer to minimize the number of parcels necessary to deliver the ordered products from the factory to the customer. A good program solving the problem of finding the minimal number of parcels necessary to deliver the given products according to an order would save a lot of money. You are asked to make such a program.

Input

The input file consists of several lines specifying orders. Each line specifies one order. Orders are described by six integers separated by one space representing successively the number of packets of individual size from the smallest size 1*1 to the biggest size 6*6. The end of the input file is indicated by the line containing six zeros.

Output

The output file contains one line for each line in the input file. This line contains the minimal number of parcels into which the order from the corresponding line of the input file can be packed. There is no line in the output file corresponding to the last ``null'' line of the input file.

Sample Input

0 0 4 0 0 1 
7 5 1 0 0 0 
0 0 0 0 0 0 

Sample Output

2 
1 

Source

Central Europe 1996

 

 

 

 

大致题意：

一个工厂制造的产品形状都是长方体盒子，它们的高度都是 h，长和宽都相等，一共有六个型号，分别为1*1, 2*2, 3*3, 4*4, 5*5, 6*6。

这些产品通常使用一个 6*6*h 的长方体箱子包装然后邮寄给客户。因为邮费很贵，所以工厂要想方设法的减小每个订单运送时的箱子数量BoxNum。

 

解题思路：

由于盒子和箱子的高均为h，因此只需考虑底面积的空间。

 

6*6的盒子，每个盒子独占一个箱子。

5*5的盒子，每个盒子放入一个箱子，该箱子的剩余空间允许放入的最大尺寸为1*1，且最多放11个。

4*4的盒子，每个盒子放入一个箱子，该箱子的剩余空间允许放入的最大尺寸为2*2。

3*3的盒子，每4个刚好独占一个箱子，不足4个3*3的，剩下空间由2*2和1*2填充。

2*2的盒子和1*1的盒子主要用于填充其他箱子的剩余空间，填充后的多余部分才开辟新箱子装填。

 

#include<iostream>
#include<cstdio>
using namespace std;  
int main()
{
	int p[4]={0,5,3,1},a[10];  //3×3的放完后，余下的放入新箱子后，还可以放几个2×2的包裹（下标对应余数）
	while(1)
	{
		int sum=0;
		for(int i=1;i<=6;i++)
		{
			scanf("%d",&a[i]);
			sum+=a[i];
		}
		if(sum==0)
			break;
		sum=0;
		sum=a[4]+a[5]+a[6]+a[3]/4;  
		if(a[3]%4!=0) sum++;
		int need2=a[4]*5+p[a[3]%4];        
		if(need2<a[2])                     
		{
			sum+=(a[2]-need2)/9;
			if((a[2]-need2)%9!=0) sum++;
		}
		int need1=sum*36-a[2]*4-a[3]*9-a[4]*16-a[5]*25-a[6]*36;   //需要的1×1的个数，即所有箱子的总面积减去后5种盒子的总面积 
		if(need1<a[1])                                            
		{
			sum+=(a[1]-need1)/36;        
		    if((a[1]-need1)%36!=0)
				sum++;
		}
		printf("%d\n",sum);
	}
	return 0;
}