二分贪心 T - 20

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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

这道题的基本题意为有1*1，2*2，3*3，4*4，5*5，6*6的产品用6*6的盒子装，要求求需要的最少盒子数。

思路为先装大的产品，6*6的只能装一个，不能装其他的；5*5的也只能装一个，还可以装11个1*1的；4*4的也只能装一个，还可以装5个2*2的，2*2的不够用1*1的补上；3*3的可以装1、2、3、4个，4个的时候刚好装满，3个的时候还能装1个2*2的跟5个1*1的，2个时候可以装3个2*2的跟6个1*1的，1个的时候可以装5个2*2的跟7个1*1的；2*2的每9个刚好装满，余下的用1*1的补装。基本贪心思路为这样。

源代码如下：

#include<iostream>
#include<cstring>
using namespace std;
int main()
{ int a[7],i,m;
 while(cin>>a[1]>>a[2]>>a[3]>>a[4]>>a[5]>>a[6])
 { 
   m=0;
  if(a[1]==0&&a[2]==0&&a[3]==0&&a[4]==0&&a[5]==0&&a[6]==0)break;
   m+=a[6];
   
   m+=a[5];
   a[1]-=a[5]*11;
   
   m+=a[4];
   a[2]-=a[4]*5;
   if(a[2]<0)
   {
     a[1]+=a[2]*4;
     a[2]=0;
   }
   
   if(a[3]%4==0)m+=a[3]/4;
   else {
         m+=a[3]/4+1;
        if(a[2]>=0)
         { if(a[3]%4==1)
		  { a[2]-=5;
		    a[1]-=7;
	      }
           if(a[3]%4==2)
		   {
		   a[2]-=3;
		   a[1]-=6;
	       }
           if(a[3]%4==3)
		   { a[2]-=1;
	         a[1]-=5;
		   }
         }
        if(a[2]<0)
		{
         a[1]+=a[2]*4;
         a[2]=0;
        }
        }
        
   if(a[2]%9==0)m+=a[2]/9;
   else 
   {
       m+=a[2]/9+1;
       a[1]-=36-(a[2]%9)*4;
    }
    
   if(a[1]>0)
    if(a[1]%36==0)m+=a[1]/36;
   else m+=a[1]/36+1;
   cout<<m<<endl;
 }
}