ACM ProblemR

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我现在做的是编号为1017的试题，具体内容如下所示：

Problem R

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 20000/10000K (Java/Other)

Total Submission(s) : 88 Accepted Submission(s) : 20

Problem 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

简单题意：有不同尺寸的箱子，包括1*1, 2*2，3*3, 4*4, 5*5, 6*6这几种规格的，把它们放在6*6规格的箱子，至少需要多少个箱子.

解题思路：

放置箱子时，优先考虑尺寸大的箱子，4*4， 5*5， 6*6，都需要占用一个箱子，对于3*3的箱子，每四个正好占有一个箱子，然后判断是否有剩余空间，

如果有的话，把2*2的箱子放到空余的空间，如果还有剩余空间的话，判断1*1的箱子能放到剩余空间多少个，如果不够的话，再根据剩余的1*1和2*2规格的

箱子，看需要多少箱子。

编写代码：

    #include<iostream>
    #include<cmath>
    #include<algorithm>
    using namespace std;

    int tab[]={0,5,3,1};
    int slove(int a[])
    {
        int num3=ceil(a[3]/4.0);
        int ans=a[6]+a[5]+a[4]+num3;
        int num2=5*a[4]+tab[a[3]%4];
        num2=min(num2,a[2]);
        a[2]-=num2;
        int rest=36*ans-a[6]*36-a[5]*25-a[4]*16-a[3]*9-num2*4;
        rest=min(rest,a[1]);
        a[1]-=rest;
        rest=a[2]*4+a[1];
        return ans+ceil(rest/36.0);
    }
    int main()
    {
        int a[10];
        while(cin >> a[1])
        {
            int kase=a[1];
            for(int i=2;i<=6;i++)
            {
                cin >> a[i];
                if(a[i])
                {
                    kase=1;
                }
            }
            if(!kase)
            {
                break;
            }
            cout << slove(a) << endl;
        }
        return 0;
    }