贪心？——Packets

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

差不多的思路是有的。
目前做的贪心题目都比较简单。
但是，为什么敲完代码迫不及待去试试样例呢？就算样例对了又有啥用呢？
为什么总是想哎呀，我的就是对的呢？
思路都有但是细节总是比别人的累赘呢？
其实这道题的题目自己还是迫不及待去看别人的讲解的。
计算多余的two都是一样的。
因为繁琐的计算one免不了出错。
一会儿发现一个真的很没意思。
其实写代码是要胸有成竹敲之前就考虑所有细节的。
其实能优化就不繁琐，多找规律套公式会比分步讨论正确率高很多的。
其实自己还是很差劲的。

#include<iostream>
using namespace std;
int a[4]={0,5,3,1};
int main(){
    int one,two,three,four,five,six;
    while(cin>>one>>two>>three>>four>>five>>six&&(one+two+three+four+five+six))
    {
        int sum=0,more2=0;
        sum+=six+five+four+three/4;
        if(three%4)
            sum++;
        more2=four*5+a[three%4];
        if(more2<two){
            sum+=(two-more2)/9;
            if((two-more2)%9)
                sum++;
        }
        int need = sum * 36 - two * 4 - three * 9 - four * 16 - five * 25 - six * 36;
        if(need<one){
             sum+=(one-need)/36;
            if((one-need)%36)
                sum++;
        }
        cout<<sum<<endl;
    }
    return 0;
}