贪心算法8之1017

1 题目编号：1017 problemR

2 题目内容：

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 

 

3 解题思路形成过程：对于6*6，5*5以及4*4尺寸的物品每个物品需要占有一个箱子，对于3*3的物品一个箱子可以放4个，2*2的物品箱子可以放9个，1*1的可以放36个。采用面积统计1*1箱子的空位，采用向上取整的方法统计箱子。

4 感想：做acm题啊，就是要有耐心，要仔细想清楚各种情况，考虑全面。

5 代码：

#include<iostream>
#include<stdlib.h>
using namespace std;
int num[4]={0,5,3,1};
int box[7];

int main(){
    while(1){
        int tmp=0;
        for(int i=1;i<=6;i++){
            cin>>box[i] ;
            tmp+=box[i];
        }
        if(tmp==0)
            break;
        int ans=box[6]+box[5]+box[4]+(box[3]+3)/4;
        int a2=box[4]*5+num[box[3]%4]; 
        if(box[2]>a2)
            ans+=(box[2]-a2+8)/9;
        int a1=ans*36-box[6]*36-box[5]*25-box[4]*16-box[3]*9-box[2]*4;
        if(box[1]>a1) 
            ans+=(box[1]-a1+35)/36;
        cout<<ans<<endl;
    }
    return 0;
}