POJ：1017-Packets（贪心+模拟，神烦）

传送门：http://poj.org/problem?id=1017

Packets

Time Limit: 1000MS Memory Limit: 10000K

Total Submissions: 59106 Accepted: 20072

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

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解题心得：

1. 这个题思维还是很简单的，那就是在之前装箱的时候要尽可能的装满，并且要从大的开始装，为啥要这样呢，如果不从大的开始装，那么到了后面，小的用完之后几个大的箱子开始装就会产生空隙，并且没有小的箱子去填满空隙。
2. 但是还有要注意的地方，这是体积，计算的时候不能按照平面来算，要用体积来计算，写起来非常的烦人，要仔细。其实只有两种装法，从大的开始和从小的开始，想一想也就知道该怎么贪心了。

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#include <stdio.h>
#include <algorithm>
using namespace std;
int box[7];
int main() {
    while(scanf("%d%d%d%d%d%d",&box[1],&box[2],&box[3],&box[4],&box[5],&box[6]) && (box[1]+box[2]+box[3]+box[4]+box[5]+box[6])) {
        int ans = 0;
        ans += box[6];

        ans += box[5];//5+1
        box[1] = max(0, box[1] - box[5] * 11);

        ans += box[4];//4+2+1
        if (box[2] < box[4] * 5) box[1] = max(0, box[1] - (5 * box[4] - box[2]));
        box[2] = max(0, box[2] - 5 * box[4]);

        ans += (box[3] + 3) / 4;//3+2+1
        box[3] %= 4;
        if (box[3] == 1) {
            if (box[2] < 5) box[1] = max(0, box[1] - (27 - 4 * box[2]));
            else box[1] = max(0, box[1] - 7);
            box[2] = max(0, box[2] - 5);
        } else if (box[3] == 2) {
            if (box[2] < 3) box[1] = max(0, box[1] - (18 - 4 * box[2]));
            else box[1] = max(0, box[1] - 6);
            box[2] = max(0, box[2] - 3);
        } else if (box[3] == 3) {
            if (box[2] < 1) box[1] = max(0, box[1] - (9 - 4 * box[2]));
            else box[1] = max(0, box[1] - 5);
            box[2] = max(0, box[2] - 1);
        }

        ans += (box[2] + 8) / 9;//2+1
        box[2] %= 9;

        if (box[2])
            box[1] = max(0, box[1] - (36 - 4 * box[2]));

        ans += (box[1] + 35) / 36;

        printf("%d\n", ans);
    }
    return 0;
}