Problem Q

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

题意：有六种规格的货物，1*1， 2*2 ，3*3， 4*4 ，5*5
，6*6装货物的箱子只有6*6的，求用的最小箱子数；

解题思路：刚开始还以为箱子都是立体的，***（自己体会）！，写到装3*3的时候就不会写了，今早上一看题，才发现昨晚自己瞎了，既然是二维的就好做了，假期看的书里有过这到题，当时看了一点，有点印象，从6*6到1*1贪心，6
，5 ，4的只能装一个，5 的还能装1的，4的还能装1 2的，3的比较麻烦能装1 2 3的，一个个的讨论，剩下1
2的，判断是否大于零就行了，因为装起来比较简单；

感悟：这道题做完比较有成就感，因为测试了很多数据，一个个的BUG改，真爽啊；

测试数据：

Input:

0 0 4 0 0 1

7 5 1 0 0 0

36 9 4 1 1 1

0 9 4 1 1 0

0 0 4 0 0 0

36 0 0 0 0 0

0 9 0 0 0 0

79 96 94 30 18 14

53 17 12 98 76 54

83 44 47 42 80 3

15 26 13 29 42 40

41 61 36 90 54 66

78 56 445 45 23 65

13 4 8 29 45 3

15 75 45 98 34 53

40 9 0 2 0 0

41 9 0 2 0 0

44 0 0 0 4 0

0 2 3 0 0 0

37 7 2 0 1 0

12 2 0 1 0 0

13 2 0 1 0 0

0 0 0 0 0 0



Output:

2

1

6

4

1

1

1

86

231

137

115

219

245

79

197

3

4

4

2

3

1

2

代码：

#include

#include

#include

using namespace std;

int main()

{

   
//freopen("in.txt", "r", stdin);

    int
ans[7]={0},sizen[7]={0,1,2,3,4,5,6};

    int
null=0;

   
while(~scanf("%d%d%d%d%d%d",&ans[1],&ans[2],&ans[3],&ans[4],&ans[5],&ans[6]))

{

       
if(ans[1]==0&&ans[2]==0&&ans[3]==0&&ans[4]==0&&ans[5]==0&&ans[6]==0)

break;

       
null=0;

       
null+=ans[6];//因为装了6*6的就没有别的空间了；

       
//装5*5的东西

       
null+=ans[5];//一个5*5的肯定需要一个箱子装

       
if(ans[1]<=11*ans[5])//还有空装1*1的盒子不要浪费

       
{

           
ans[1]=0;

       
}

       
else

       
{

           
ans[1]-=11*ans[5];

       
}

       
//printf("null=%d\n",null);

       
//printf("a[1]=%d a[2]=%d a[3]=%d a[4]=%d a[5]=%d
a[6]=%d\n",ans[1],ans[2],ans[3],ans[4],ans[5],ans[6]);

       
//装4*4的东西

       
null+=ans[4];//一个4*4的肯定需要一个箱子装,余下的能成5个2*2

       
//或20个1*1的

       
int content=20*ans[4];//这是剩余的盒子空间；

       
if(ans[2]<=5*ans[4])//先装2*2的盒子

       
{

           
int content=20*ans[4];//这是剩余的盒子空间；

           
content-=ans[2]*4;//剩下的体积

           
if(ans[1]<=content)//现在装1*1的盒子

           
{

               
ans[1]=0;

           
}

           
else

           
{

               
ans[1]-=content;

           
}//剩下几平方米就能装几个1*1的；

           
ans[2]=0;

       
}

       
else

       
{

           
ans[2]-=5*ans[4];

       
}

       
ans[4]=0;

       
//printf("null=%d\n",null);

       
//printf("a[1]=%d a[2]=%d a[3]=%d a[4]=%d a[5]=%d
a[6]=%d\n",ans[1],ans[2],ans[3],ans[4],ans[5],ans[6]);

       
//装3*3的

       
null+=ans[3]/4;

       
ans[3]=ans[3]%4;//每四个3*3能装满一个箱子；

       
if(ans[3]>0)

       
{

           
null+=1;

           
if(ans[2]<=7-ans[3]*2)//先装2*2的盒子

           
{

               
int content=36-ans[3]*9-ans[2]*4;//这是剩余的盒子空间；

               
if(ans[1]<=content)//现在装1*1的盒子

               
{

                   
ans[1]=0;

               
}

               
else

               
{

                   
ans[1]-=content;

               
}//剩下几平方米就能装几个1*1的；

               
ans[2]=0;

           
}

           
else

           
{

               
int content=36-ans[3]*9-(7-ans[3]*2)*4;

               
if(ans[1]<=content)//现在装1*1的盒子

               
{

                   
ans[1]=0;

               
}

               
else

               
{

                   
ans[1]-=content;

               
}//剩下几平方米就能装几个1*1的；

               
ans[2]-=(7-ans[3]*2);

           
}

           
ans[3]=0;

           
//printf("null=%d\n",null);

           
//printf("a[1]=%d a[2]=%d a[3]=%d a[4]=%d a[5]=%d
a[6]=%d\n",ans[1],ans[2],ans[3],ans[4],ans[5],ans[6]);

       
}

       
//装2*2的

       
if(ans[2]>0)

       
{

           
null+=ans[2]/9;

           
ans[2]%=9;

           
if(ans[2]>0)

           
{

               
null+=1;

               
int content=36-ans[2]*4;

               
if(ans[1]<=content)//现在装1*1的盒子

               
{

                   
ans[1]=0;

               
}

               
else

               
{

                   
ans[1]-=content;

               
}//剩下几平方米就能装几个1*1的；

           
}

           
ans[2]=0;

           
//printf("null=%d\n",null);

           
//printf("a[1]=%d a[2]=%d a[3]=%d a[4]=%d a[5]=%d
a[6]=%d\n",ans[1],ans[2],ans[3],ans[4],ans[5],ans[6]);

       
}

       
//装1*1的

       
if(ans[1]>0)

       
{

           
null+=ans[1]/36;

           
ans[1]%=36;

           
if(ans[1])

               
null+=1;

           
ans[1]=0;

           
//printf("null=%d\n",null);

           
//printf("a[1]=%d a[2]=%d a[3]=%d a[4]=%d a[5]=%d
a[6]=%d\n",ans[1],ans[2],ans[3],ans[4],ans[5],ans[6]);

       
}

       
printf("%d\n",null);

    }

}