poj1017

一、题意：有底面积为1*1,2*2，...，6*6高为h的物品若干，求用最少的包装盒包装好这些物品，其中包装盒的大小固定为底面积6*6，高为h

二、思路：贪心。先放大的，然后用小的尽量去凑满这个包装盒，从6到1类推。我用的是比较暴力的方法，每一种情况都考虑出来，然后模拟。

三、代码：

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#include"iostream" #include"stdio.h" #include"math.h" using namespace std;   int num[7];   int Solve() {     int res=num[5];     res+=num[4];     num[0]-=num[4]*11;     res+=num[3];     if(num[1]<num[3]*5)     {         num[0]-=(20*num[3]-4*num[1]);     }     num[1]-=num[3]*5;     res+=num[2]/4;     int numLeave=num[2]%4;     if(numLeave>0) res++;     switch(numLeave)     {     case 1:         if(num[1]<5)         {             if(num[1]>0)             {                 num[0]-=(27-num[1]*4);                 num[1]=0;             }             else                 num[0]-=27;           }         else         {             num[1]-=5;             num[0]-=7;         }         break;     case 2:         if(num[1]<3)         {             if(num[1]>0)             {                 num[1]=0;                 num[0]-=(18-num[1]*4);             }             else                 num[0]-=18;         }         else         {             num[1]-=3;             num[0]-=6;         }         break;     case 3:         if(num[1]<1)         {             num[0]-=9;         }         else         {             num[1]-=1;             num[0]-=5;         }         break;     default:         break;     }     if(num[1]>0)     {         res+=num[1]/9;         if(num[1]%9>0)         {             res++;             num[0]-=(36-4*(num[1]%9));         }     }     if(num[0]>0)     {         res+=ceil(num[0]/36.0);     }     return res; }   int main() {     while(true)     {         int sum=0;         for(int i=0;i<6;i++)         {             cin>>num[i];             sum+=num[i];         }         if(!sum) break;         cout<<Solve()<<endl;     }     return 0; }