poj 1017 -- Packets

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 

 1 #include<iostream>

 2 using namespace std;

 3 

 4 int main() {

 5     int N, a, b, c, d, e, f, y, x;//N储存箱子数，y储存2*2的空位数，

 6                                   //x储存1*1的空位数

 7     int u[4]={0,5,3,1};//数组u表示3*3的产品数目分别是四的倍数，四的倍数+1,四的倍数+2,四的倍数+3 时 ，对于2*2的空位数；

 8     while(1) {

 9         cin >> a >> b >> c >> d >> e >> f;

10         if(a==0 && b==0 && c==0 && d==0 && e==0 && f==0)break;

11         N = f+e+d+(c+3)/4;//(c+3)/4等于c除以4向上去整

12         y = 5*d + u[c%4];

13         if(b > y) N += (b-y+8)/9;

14         x = 36*N-36*f-25*e-16*d-9*c-4*b;

15         if(a>x) N += (a-x+35)/36;

16         cout << N << "\n";

17     }

18     return 0;

19 }

题目比较好理解。有一个关键点，就是先考虑2*2的，再考虑1*1的。

而且要善用空余数量。

（这题用贪心可能就会写的比较长）