Packets-贪心

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 

题意:工厂生产大小分别为1*1,2*2,3*3,4*4,5*5,6*6的产品,产品高为h,现在要把这些产品装入6*6,高也为h的包裹里,问至少需要多少包裹.

输入的6个数字分别表示1*1,2*2,3*3,4*4,5*5,6*6产品的个数(代码里用abcdef表示).

思路:先装大的,再装小的,在装3*3或4*4时,要把2*2的插进去,最后面积剩下的再把1*1插进去

1个包裹最多能按下面的装:

1个6*6

1个5*5

1个4*4+5个2*2

4个3*3

3个3*3+1个2*2

2个3*3+3个2*2

1个3*3+5个2*2

9个2*2

36个1*1

#include<stdio.h>
#include<string.h>
#include<string>
#include<algorithm>
#include<iostream>
using namespace std;
int main()
{
    int a,b,c,d,e,f,i,cnt,x,b1;
    while(scanf("%d%d%d%d%d%d",&a,&b,&c,&d,&e,&f)!=EOF&&(a!=0||b!=0||c!=0||d!=0||e!=0||f!=0))
    {
        b1=b;
        cnt=f+e;
        if(d>0)
        {
            cnt+=d;
            b-=d*5;
        }
        if(c>0)
        {
            cnt+=c/4;
            x=c%4;
            if(x>0)
            cnt++;
            if(x==3)
            b-=1;
            else if(x==2)
            b-=3;
            else if(x==1)
            b-=5;
        }
        if(b>0)
        {
            cnt+=b/9;
            x=b%9;
            if(x>0)
            cnt++;
        }
        a=a-(cnt*36-f*36-e*25-d*16-c*9-b1*4);
        if(a>0)
        {
            cnt+=a/36;
            x=a%36;
            if(x>0)
            cnt++;
        }
        printf("%d\n",cnt);
    }
}