内存:128  时间:1

题目描述

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.

输入

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.

输出

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.

样例输入

0 0 4 0 0 1
7 5 1 0 0 0
0 0 0 0 0 0

样例输出

2
1

提示

代码如下

#include<iostream>
#include<cstdio>
#include<cstring>

int num[4]={0,5,3,1};
int box[7];

int main(){

    //freopen("input.txt","r",stdin);

    while(1){
        int tmp=0;
        for(int i=1;i<=6;i++){
            scanf("%d",&box[i]);    // 读入每个物品的数目
            tmp+=box[i];
        }
        if(tmp==0)
            break;
        int ans=box[6]+box[5]+box[4]+(box[3]+3)/4;  //a6,a5,a4,每个物品占有一个箱子(a3 + 3 ) / 4 代表a3的物品需要占
        int a2=box[4]*5+num[box[3]%4];  //统计所有的大物品放进箱子中后a2物品的空位子有多少
        if(box[2]>a2)
            ans+=(box[2]-a2+8)/9;
        int a1=ans*36-box[6]*36-box[5]*25-box[4]*16-box[3]*9-box[2]*4;
        if(box[1]>a1)   //求a1的空位子,只需要统计剩余的面积即可
            ans+=(box[1]-a1+35)/36;
        printf("%d
",ans);
    }
    return 0;
}

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