POJ 1017 Packets 贪心

Packets

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 53481 Accepted: 18185

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 

Source

Central Europe 1996

题目大意是有六种型号的包裹，试问用几个6*6的大包裹可以将给定的包裹放在里面。

贪心的思路就是从型号最大的开始枚举，6，5，4三种型号分别需要占用一个大包裹。3号讨论一下剩余的情况（每4个3号的占用一个大包裹）。之后再放2，1型号的。

五组数据，仅供参考

0 0 4 0 0 1 7 5 1 0 0 0 8 5 1 0 0 065 65 513 1 8 858 63 33 3 3 321215324

#include <iostream>#include <cstdio>using namespace std;int main(){    int a, b, c, d, e, f;    int no[4] = {0, 5, 3, 1};    while(scanf("%d%d%d%d%d%d", &a, &b, &c, &d, &e, &f) != EOF) {        if(a == 0 && b == 0 && c == 0 && d == 0 && e == 0 && f == 0) {            break;        }        int ans = f + e + d + (c + 3) / 4;  //目前的大包裹数        int temp = 5 * d + no[c % 4];       //2号包裹可用的空余位置        if(b >= temp) {            ans += (b - temp + 8) / 9;              }        temp = ans * 36 - f * 36 - e * 25 - d * 16 - c * 9 - b * 4; //1号包裹可用的空余位置        if(a >= temp) {            ans += (a - temp + 35) / 36;        }        printf("%d\n", ans);    }    return 0;}