1017 of greedy strategy*
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
题目要求：每个包裹都是6*6的，而物件有1*1，2*2..........6*6。如何分配物件使所用的包裹数最小。
思路：先用一个数组保存各类物品的数量，然后进行从大到小判断。6*6:直接计数；5*5：直接计数，并且1*1的减11个；4*4：直接计数，其数目*5与2*2的数目比较，若大于则2*2清0，1*1的减去其差值乘4，否者2*2的减少4*4数目的五倍;3*3:数目除以4后计数，并将余数分为4种情况。没有余数跳过，有余数计数加一。若余数为3，则判断2*2的目是否大于0，大于则2*2减1，1*1的减5，若余数为2，则判断2*2的是否大于3，大于减3，1*1的减6，否者2*2清零并1*1减取3与2*2差值*4+6，若余数为一，则判断5和2*2的数目，若2*2多则其数目减5，1*1的减去6，否者2*2清零，1*1的减去5与2*2的差值*4+6；2*2：若不为0，则计数加上其数目/9,若模9有余数则再加一，1*1的数目减去36和其余数*4的差。1*1:若不小于1则计数加其数目/36，有余数计数再加一。
细节：题目思路没难度，只是复杂，要细心写判断语句。
#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;
}
}