poj 1017 Packets 模拟题
来源:互联网 发布:文明网络行教学反思 编辑:程序博客网 时间:2024/06/08 03:33
Packets
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 41736 Accepted: 14018
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
本题大意为把6种不同尺寸(分别是1*1,2*2,3*3,4*4,5*5,6*6) 不同数量的产品用6*6的包装盒来包装,求最小包装盒数。主要关键在于弄清题意,然后先包装尺寸大的,其他剩余空间用尺寸小的产品补充,尤其注意讨论3*3尺寸包装的剩余空间补足情况。
#include <cstdio>//本题主要考察模拟包装物品操作int square[7]; //表示目前包装盒剩余空间对应的size数量int parcel[7]; //表示输入的产品尺寸对应的数量int main(){ bool flag = true; int num = 0; while(1) { flag = true; for(int i = 1; i <= 6 ; i++) { scanf("%d", &parcel[i]); if(parcel[i] != 0) flag = false; } if(flag) break; num = parcel[6] + parcel[5] + parcel[4]; //包装尺寸为6*6和5*5和4*4的只能增加包装盒的数量 num = num + parcel[3] / 4; square[3] = 0; if(parcel[3] % 4 != 0) //假如尺寸为3*3的包装没有装满包装盒 { num ++; square[3] = 4 - (parcel[3] % 4); } square[1] = 11 * parcel[5]; //包装5*5产品所剩下的1*1尺寸个数,可供1*1产品使用 square[2] = 5 * parcel[4];//包装4*4产品所剩2*2尺寸个数,可供2*2和1*1使用 if(square[3] == 1 || square[3] == 0) { square[2] = square[2] + square[3];//剩下的能容纳2*2的空间 square[1] = square[1] + square[3] * 5;//目前只能容纳1*1的空间数量 } else if(square[3] == 2) { square[2] = square[2] + 3; square[1] = square[1] + 6; } else { square[2] = square[2] + 5; square[1] = square[1] + 7; } if(parcel[2] <= square[2]) square[1] = square[1] + 4 * (square[2] - parcel[2]); else { parcel[2] = parcel[2] - square[2]; num = num + parcel[2] / 9; if(parcel[2] % 9 != 0) { num++; square[1] += 4 * (9 - (parcel[2] % 9)); } } if(square[1] < parcel[1]) { parcel[1] = parcel[1] - square[1]; num = num + parcel[1] / 36; if(parcel[1] % 36 != 0) num = num + 1; } printf("%d\n", num); }return 0;}
0 0
- poj 1017 Packets 模拟题
- POJ 1017 Packets ----模拟
- POJ 1017 Packets 贪心 + 模拟
- 贪心+模拟-poj-1017-Packets
- POJ 1017 Packets 模拟法
- POJ 1017 Packets(模拟)
- POJ 1017Packets---简单模拟
- POJ 1017 Packets 简单模拟
- POJ 1017 Packets 【贪心 模拟】
- POJ 1017 Packets 悼念死去的模拟
- POJ 1017 Packets(构造模拟)
- POJ 1017 - Packets(贪心+模拟)
- POJ 1017 Packets(贪心,好题)
- poj算法题1017——Packets
- poj 1017 Packets【贪心】
- POJ 1017 Packets(贪心)
- poj 1017 Packets
- POJ 1017 Packets
- 听说是大数阶乘,没测过!!
- Inside the c++ object model读书笔记之拷贝构造函数(二)
- ASP.NET学习笔记(五)-全球化部署,网站发布方法,AJAX使用,水晶报表使用,DropDownList,CheckBox全选
- tlplayer,wzplayer支持wince,winphone,windows8 for arm
- Android 用剪切板传递数据
- poj 1017 Packets 模拟题
- Window,Document,Frame的理解,帮助理解左侧导航栏的实现原理
- mysql source 防止乱码
- Django学习笔记(一)-入门
- Spring学习笔记--AOP详解
- Linux操作系统下 MySQL的服务器字符集设置
- LeetCode之Reorder List
- eclipse中jsp代码 request response等不提示问题解决方案
- css