# 贪心算法(各种贪心题目)

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 题目大意：有1*1,2*2···6*6尺寸的箱子，每种物品的高度确定，只用6*6的箱子，输入每种尺寸的物品有多少个，问你最少要用多少箱子，当六个0时推出循环个人思路：6*6，5*5，4*4的物品每一个都需要一个箱子来放，所以先考虑，4个3*3物品占满一个箱子，考虑要多少个箱子，然后考虑已经用了的箱子能放多少个2*2的，不够的话再用箱子最后考虑总共用了多少个箱子，把方块数求出来，减去用掉的方块数，就算当前的箱子能放的1*1的物品数量，不够再补箱子```
```#include<iostream>
#include<stdio.h>
#include<string.h>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<set>
typedef long long ll;
using namespace std;
const ll mod=1e9+7;
#define INF 0x3f3f3f
int main()
{
ll a[10];
ll num[4]={0,5,3,1};
while(1)
{
ll ans=0;
ll tmp=0;
for(int i=1;i<=6;i++)
{
scanf("%I64d",&a[i]);
tmp+=a[i];
}
if(tmp==0)
return 0;
ans+=a[6]+a[5]+a[4]+(a[3]+3)/4;//a6,a5,a4每个物品需要一个箱子，4个a3需要一个箱子
ll a2=a[4]*5+num[a[3]%4];//a6,a5,a4,a3有多少个空位子可以放2*2的物品
if(a[2]>a2)
{
ans+=(a[2]-a2+8)/9;
}
int a1=ans*36-a[6]*36-a[5]*25-a[4]*16-a[3]*9-a[2]*4;//总共用的箱子还剩多少个能放a1
if(a[1]>a1)
{
ans+=(a[1]-a1+35)/36;
}
printf("%I64d\n",ans);
}
return 0;
}
```

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