ACM1017箱子问题

问题描述：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

意思：有一批箱子1*1,2*2,3*3...6*6，将这批箱子装到6*6的大箱子里，求出所需大箱子最少的个数。输出为六个箱子的个数。

思路：6*6,5*5,4*4必定一个需要一个6*6的大箱子，4个3*3的箱子需要一个大箱子，2*2的箱子和4*4,3*3的相对。

代码：

#include <iostream>
#include<string.h>
using namespace std;

int box[7];
int sum[4] = {0,5,3,1};//对应3*3剩余空间可以放的2*2箱子
bool input()
{
　　int flag = 0;
　　memset(box,0,sizeof(box));
　　for(int i = 1;i < 7;i++)
　　{
　　　　cin >> box[i];
　　　　flag += box[i];
　　}
　　return flag;
}
int main()
{
　　int ans = 0;
　　int num = 0;
　　num = input();
　　while(num)
　　{
　　　　ans = box[6] + box[5] + box[4] + (box[3]+3)/4;//算出3*3,4*4，5*5,6*6所需要的箱子的个数考虑到3*3的箱子的个数可能为0，故+3

　　　　int b2 = box[4] * 5 + sum[box[3]%4];//求剩余空间可以装多少2*2的箱子
　　　　if(box[2] > b2)
　　　　{
　　　　　　ans+=(box[2]-b2+8)/9;
　　　　}
　　　　int b1=ans*36-box[6]*36-box[5]*25-box[4]*16-box[3]*9-box[2]*4;
　　　　if(box[1]>b1)
　　　　　　ans+=(box[1]-b1+35)/36;
　　　　cout << ans << endl;
　　　　num = input();
　　}
　　return 0;
}

 

 

 

 

转载于:https://www.cnblogs.com/2016zhanggang/p/5339951.html