joj1538

 1538: Packets

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	311	108	Standard

A factory produces products packed in square packets of the same height h and of the sizes ,  ,  ,  ,  ,  . These products are always delivered to customers in the square parcels of the same height h as the products have and of the size  . 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  to the biggest size  . 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

#include<stdio.h>
int a[7];
int main()
{
    freopen("in.txt","r",stdin);
    while(scanf("%d%d%d%d%d%d",&a[1],&a[2],&a[3],&a[4],&a[5],&a[6])==6)
    {
        if(a[1]+a[2]+a[3]+a[4]+a[5]+a[6]==0)   break;
        int count=0;
        int temp;
        count+=a[6]+a[5]+a[4];
        int left3=0;
        if(a[3]%4==0)
        {
            count+=a[3]/4;
            left3=0;
        }
        else
        {
            count+=a[3]/4+1;
            left3+=(4-a[3]%4);
        }
        int left2=a[4]*5;
        int left1=a[5]*11;
        if(left3==3)
        {
            left2+=5;
            left1+=7;
        }
        if(left3==2)
        {
            left2+=3;
            left1+=6;
        }
        if(left3==1)
        {
            left2+=1;
            left1+=5;
        }
        if(a[2]>left2)
        {
            a[2]-=left2;
            if(a[2]%9==0)
            count+=a[2]/9;
            else
            {
                count+=a[2]/9+1;
                left1+=4*(9-a[2]%9);
            }
        }
        else
        {
           left2-=a[2];
           left1+=left2*4;
        }
        if(a[1]>left1)
        {
            a[1]-=left1;
            if(a[1]%36==0)
            count+=a[1]/36;
            else
            count+=a[1]/36+1;
        }
        printf("%d\n",count);
    }
    return 0;
}
对于这道题目我是感触颇深，这个题目一开始我就想错了，以为可以将整个6*6的
方格进行划分结果弄了一个小时也没有调试出来，最后是换了一种思路才调试出来
下面是我第一次的错误代码。对于1个3*3和5个2*2的题目下面的代码将是错误的。

/*#include<stdio.h>
int packet[7];
bool exam()
{
    for(int i=1;i<=6;i++)
    {
        if(packet[i]!=0)
        return false;
    }
    return true;
}
void cal(int u,int v)
{
    for(int i=6;i>0;i--)
    {
        if(packet[i]!=0)
        {
            if(u>=i&&v>=i)
            {
                packet[i]--;
                if(u>v)
                {
                    if(u-i>v)
                        cal(u-i,v);
                    else
                        cal(v,u-i);
                    if(i>v-i)
                        cal(i,v-i);
                    else
                        cal(v-i,i);
                    v=v-i;
                    u=u-i;
                }
                else
                {
                    if(i>u-i)
                        cal(i,u-i);
                    else
                        cal(u-i,i);
                    if(v-i>u)
                        cal(v-i,u);
                    else
                        cal(u,v-i);
                    v=v-i;
                    u=u-i;
                }
            }
        }
    }
}


int main()
{
    freopen("in.txt","r",stdin);
   // freopen("out.txt","w",stdout);
    bool first=true;
    while(1)
    {
        bool temp=false;
        for(int i=1;i<=6;i++)
        {
            scanf("%d",&packet[i]);
            if(packet[i]!=0)
            temp=true;
        }
        if(!temp)break;
        if(!first)
        printf("\n");
        first=false;
        int num=0;
        while(!exam())
        {
            cal(6,6);
            num++;
        }
        printf("%d",num);
    }
    return 0;
}
*/