SDAU 课程练习 1017

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

题目大意：

输入包含 1 * 1   2*2  3*3  4*4  5*5  6*6  商品箱子的个数，现将全部箱子放入 6*6 的大包装箱子中，求所用最少包装箱。

思路：

6*6 的商品箱子单独占一个包装箱  5*5  的商品箱子只能用 1*1 的来补剩余空间，以此类推。所以从 6*6  到 1*1  讨论即可。（暴力解法）。

题目感想：

这个是和李跃威一起写的，因为我们思路一致，然后我敲 if else  语句，他在一旁检查，错误和卡顿的时候提醒我。200 行 if else 语句 一次写下来，没再加任何修改，一遍就AC了。后续应该会更新贪心的方法，

AC代码：​​​​​

#include <cstdio>#include<iostream>#include<stdio.h>#include<vector>#include<algorithm>#include<numeric>#include<math.h>#include<string.h>#include<map>#include<set>#include<vector>using namespace std;int main(){    //freopen("r.txt", "r", stdin);    int a[10];    int i,b,num,last;    while(cin>>a[1])    {        int count=0;        for(i=2;i<=6;i++)            cin>>a[i];        for(i=1;i<=6;i++)            if(a[i]==0) count++;        if(count==6) break;        num=a[6];        if(a[5]>0)        {            num+=a[5];            if((a[5]*11)>=a[1])            {                a[1]=0;            }            else            {                a[1]-=a[5]*11;            }        }        if(a[4]>0)        {            num+=a[4];            if(a[4]*5>=a[2])            {                last=36*a[4]-a[4]*16-a[2]*4;                a[2]=0;                if(a[1]>0)                {                    if(last>=a[1])                    {                        a[1]=0;                    }                    else                    {                        a[1]-=last;                    }                }            }            else            {                a[2]-=a[4]*5;            }        }        if(a[3]>0)        {            num+=a[3]/4;            a[3]%=4;            if(a[3]!=0)            {                num++;                last=36-a[3]*9;                if(a[3]==3)                {                    if(a[2]!=0)                    {                        a[2]--;                        if(a[1]>5)                        {                            a[1]-=5;                        }                        else                        {                            a[1]=0;                        }                    }                    else                    {                        if(a[1]>9)                        {                            a[1]-=9;                        }                        else                        {                            a[1]=0;                        }                    }                }                if(a[3]==2)                {                    if(a[2]>3)                    {                        a[2]-=3;                        if(a[1]>6)                        {                            a[1]-=6;                        }                        else                        {                            a[1]=0;                        }                    }                    else                    {                        last=18-a[2]*4;                        a[2]=0;                        if(a[1]>last)                        {                            a[1]-=last;                        }                        else                        {                            a[1]=0;                        }                    }                }                if(a[3]==1)                {                    if(a[2]>5)                    {                        a[2]-=5;                        if(a[1]>7)                        {                            a[1]-=7;                        }                        else                        {                            a[1]=0;                        }                    }                    else                    {                        last=36-9-a[2]*4;                        a[2]=0;                        if(a[1]>last)                        {                            a[1]-=last;                        }                        else                        {                            a[1]=0;                        }                    }                }            }        }        if(a[2]>0)        {            num+=a[2]/9;            a[2]%=9;            if(a[2]!=0)            {                num++;                last=36-a[2]*4;                if(a[1]>last)                {                    a[1]-=last;                }                else                {                    a[1]=0;                }            }        }        if(a[1]>0)        {            num+=a[1]/36;            a[1]%=36;            if(a[1]!=0)            {                num++;            }        }        cout<<num<<endl;    }}