POJ 1017 Packets

最新推荐文章于 2018-02-09 17:00:34 发布

BuringCode

最新推荐文章于 2018-02-09 17:00:34 发布

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本文介绍了一种用于解决如何将不同尺寸的小型方形容器最优化装入统一规格的大方形容器中的算法。该算法考虑了从最小号1*1到最大号6*6的方形产品包装，并以最小化所需大容器数量为目标。

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Packets

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 51779		Accepted: 17578

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

题目大意：输入高相同，底面积分别为1*1 2*2 3*3 4*4 5*5 6*6的盒子个数现在有底面积6*6的箱子求把输入的所有盒子都装箱最少要用多少箱子

#include <iostream>
using namespace std;
int main()
{
	std::cin.tie(false);
	ios::sync_with_stdio(false);
	int num[4] = { 0, 5, 3, 1 }; 
	/*
	如果箱子里没有放3*3的盒子则这个箱子暂时不放2*2的盒子
	如果放了一个3*3的盒子 则还可以放五个2*2的盒子
	如果放了两个3*3的盒子 则好可以放三个2*2的盒子
	如果放了三个3*3的盒子 则还可以放一个2*2的盒子
     */                         
	int a[10];
	int ans;
	while (1)
	{
	int sum=0;
		ans = 0;
		for (int i = 1; i <= 6; i++)
		{
			cin >> a[i];
			sum+=a[i];
		}
		if(sum==0) break;
			ans += a[4] + a[5] + a[6] + (a[3] + 3) / 4;  //每一个4*4 5*5 6*6的盒子都要占用一个箱子 
			int box1, box2;  //储存现在还能装入的1*1 2*2的盒子数
			box2 = a[4] * 5 + num[a[3] % 4];
			if (a[2] > box2)
			{
				ans += (a[2] - box2 + 8) / 9;
			}
			box1 = ans * 36 - a[6] * 36 - a[5] * 25 - a[4] * 16 - a[3] * 9 - a[2] * 4;
			if (a[1] > box1)
			{
				ans += (a[1] - box1 + 35) / 36;
			}
		cout << ans << endl;
	}
}