贪心 —— POJ 1017 Packets

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Packets

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 46584		Accepted: 15752

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 

题意：

装箱问题  

    问题描述 
    一个工厂制造的产品形状都是长方体，它们的高度都是h，长和宽都相等，一共有六个 
型号，他们的长宽分别为 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       

的长方体包裹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 
包裹数量。他们很需要有一个好的程序帮他们解决这个问题从而节省费用。现在这个程序由 
你来设计。  
    输入数据  
    输入文件包括几行，每一行代表一个订单。每个订单里的一行包括六个整数，中间用空 
格隔开，分别为 1*1 至6*6 这六种产品的数量。输入文件将以 6 个0 组成的一行结尾。  
    输出要求  
    除了输入的最后一行6 个0 以外，输入文件里每一行对应着输出文件的一行，每一行输 
出一个整数代表对应的订单所需的最小包裹数。  
    输入样例 
    0 0 4 0 0 1   
    7 5 1 0 0 0   
    0 0 0 0 0 0   
      
    输出样例  
    2   
    1  

思路：

高度无需考虑，只考虑面积就行。假设输入a1, a2, a3, a4, a5, a6

从面积最大的开始处理，可知，对于6*6大小的箱子，6*6的物品会霸占一个箱子；5*5的物品会余下11个能装1*1物品的空间；4*4的物品会余下20个1*1的空间，可以放2*2和1*1的物品；对于3*3的物品，所需的箱子数就是(a3 + 3) / 4（即是a3除以4向上取整），然后余数为0就刚好放完；为1就剩下27个1*1的空间，为2就剩下18个1*1的空间，为3就剩下9个1*1的空间；然后把2*2和1*1的物品尽可能多地往已用的箱子里面塞；多出的向上取整。

这题用到几个技巧可以缩短代码量，做完后观看网上的代码，好短~

本人代码：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>

int main()
{
	int a1, a2, a3, a4, a5, a6;
	int i, ans;
	int b, r, c, p, all;
	//freopen("in.txt", "r", stdin);
	while(scanf("%d%d%d%d%d%d", &a1, &a2, &a3, &a4, &a5, &a6) == 6)
	{
		if(0 == a1 + a2 + a3 + a4 + a5 + a6) break;
		ans = a6 + a5 + a4;

		c = 0;
		b = 5 * a4;
		ans += (a3 + 3) / 4;
		r = a3 % 4;
		if(1 == r) c = 5;
		else if(2 == r) c = 3;
		else if(3 == r) c = 1;
		b += c;

		p = 0;
		if(r) p = 36 - 9 * r;
		all = 11 * a5 + 20 * a4 + p;

		//if(b > a2){all -= 4 * (b - a2); a2 = 0;}
		if(b > a2){all -= 4 * a2; a2 = 0;}
		else{a2 -= b; all -= 4 * b;}

		if(all > a1) a1 = 0;
		else a1 -= all;

		ans += (a2 + 8) / 9;
		r = a2 % 9;
		all = 36 - 4 * r;
		if(r){
			if(all > a1) a1 = 0;
			else a1 -= all;
		}

		ans += (a1 + 35) / 36;

		printf("%d\n", ans);
	}
	return 0;
}

模仿网上代码：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>

int main()
{
	//freopen("in.txt", "r", stdin);
	int a1, a2, a3, a4, a5, a6;
	int b, x, ans;
	int r[] = {0, 5, 3, 1};
	while(scanf("%d%d%d%d%d%d", &a1, &a2, &a3, &a4, &a5, &a6) == 6)
	{
		if(0 == a1 + a2 + a3 + a4 + a5 + a6) break;

		ans = a6 + a5 + a4 + (a3 + 3) / 4; //4*4,5*5,6*6都可以占一个箱子

		b = 5 * a4 + r[a3%4]; //计算目前为止所开的箱子能够装下2*2的物品的最大数量

		if(a2 > b) ans += (a2 - b + 8) / 9; //2*2的物品大于最大数量，应为剩下的开新的箱子
		
		x = 36 * ans - 36 * a6 - 25 * a5 - 16 * a4 - 9 * a3 - 4 * a2; //计算目前为止所开的箱子能够装下1*1的物品的最大数量

		if(a1 > x) ans += (a1 - x + 35) / 36; //1*1的物品大于最大数量，应为剩下的开新的箱子;

		printf("%d\n", ans);
	}
	return 0;
}