Time Limit: 1000MS Memory Limit: 10000KB 64bit IO Format: %I64d & %I64u

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


思路:题目说每件物品的高都是h,那么就只用考虑长和宽,问题由三维简化为二维。
如何保证用的箱子最少?先照顾大的,然后用小的填空隙。
怎么填?
6×6:一个箱子
5×5:一个箱子,空隙用1×1的填
4×4:一个箱子,空隙用2×2和1×1的填
3×3:四个装一箱,剩下的装箱后用2×2和1×1的填
2×2,1×1:优先用来填大箱子,剩下的单独装箱
接着就是模拟填充的过程,我智商低,代码写得繁琐,网上有大神仅用20+行的代码便完成了Orz
AC代码:
#include <cstdio>
#include <algorithm>
using namespace std;
int size[10];
int ans;
bool input()
{
	bool flag = 0;
	for (int i = 1; i <= 6; i++)
	{
		scanf("%d", &size[i]);
		if (size[i])
			flag = 1;
	}
	return flag;
}
void solve()
{
	ans = 0;
	int used;
	if (size[6])
		ans += size[6];
	if (size[5])
	{
		ans += size[5];
		size[1] -= size[5] * 11;
		if (size[1] < 0)
			size[1] = 0;
	}
	if (size[4])
	{
		ans += size[4];
		used = min(size[2], 5 * size[4]);
		size[2] -= size[4] * 5;
		size[1] -= (size[4] * 20 - used * 4);
		if (size[2] < 0)
			size[2] = 0;
		if (size[1] < 0)
			size[1] = 0;
	}
	if (size[3])
	{
		ans += size[3] / 4;
		size[3] %= 4;
		if (size[3])
		{
			ans++;
			switch (size[3])
			{
				case 1:
					used = min(size[2], 5);
					size[2] -= 5;
					size[1] -= (27 - used * 4);
					if (size[2] < 0)
						size[2] = 0;
					if (size[1] < 0)
						size[1] = 0;
					break;
				case 2:
					used = min(size[2], 3);
					size[2] -= 3;
					size[1] -= (18 - used * 4);
					if (size[2] < 0)
						size[2] = 0;
					if (size[1] < 0)
						size[1] = 0;
					break;
				case 3:
					used = min(size[2], 1);
					size[2] -= 1;
					size[1] -= (9 - used * 4);
					if (size[2] < 0)
						size[2] = 0;
					if (size[1] < 0)
						size[1] = 0;
					break;
			}
		}
	}
	if (size[2])
	{
		ans += size[2] / 9;
		size[2] %= 9;
		if (size[2])
		{
			ans++;
			size[1] -= (36 - size[2] * 4);
		}
		if (size[2] < 0)
			size[2] = 0;
		if (size[1] < 0)
			size[1] = 0;
	}
	if (size[1])
	{
		ans += size[1] / 36;
		size[1] %= 36;
		if (size[1])
			ans++;
	}
	printf("%d\n", ans);
}
int main()
{
	while (input())
		solve();
	return 0;
}

 

分类: ACM/ICPC

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