Booklet Printing[HDU1117]
Booklet Printing
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 503 Accepted Submission(s): 269
Problem Description
When printing out a document, normally the first page is printed first, then the second, then the third, and so on until the end. However, when creating a fold-over booklet, the order of printing must be altered. A fold-over booklet has four pages per sheet, with two on the front and two on the back. When you stack all the sheets in order, then fold the booklet in half, the pages appear in the correct order as in a regular book.
For example, a 4-page booklet would print on 1 sheet of paper: the front will contain page 4 then page 1, and the back will contain page 2 then page 3.
Front Back
------------- -------------
| | | | | |
| 4 | 1 | | 2 | 3 |
| | | | | |
------------- -------------
Your task is to write a program that takes as input the number of pages to be printed, then generates the printing order.
Input
The input contains one or more test cases, followed by a line containing the number 0 that indicates the end of the file.
Each test case consists of a positive integer n on a line by itself, where n is the number of pages to be printed; n will not exceed 100.
Output
For each test case, output a report indicating which pages should be printed on each sheet, exactly as shown in the example. If the desired number of pages does not completely fill up a sheet, then print the word Blank in place of a number. If the front or back of a sheet is entirely blank, do not generate output for that side of the sheet.
Output must be in ascending order by sheet, front first, then back.
Sample Input
1
14
4
0
Sample Output
Printing order for 1 pages:
Sheet 1, front: Blank, 1
Printing order for 14 pages:
Sheet 1, front: Blank, 1
Sheet 1, back : 2, Blank
Sheet 2, front: 14, 3
Sheet 2, back : 4, 13
Sheet 3, front: 12, 5
Sheet 3, back : 6, 11
Sheet 4, front: 10, 7
Sheet 4, back : 8, 9
Printing order for 4 pages:
Sheet 1, front: 4, 1
Sheet 1, back : 2, 3
Source
Mid-Central USA 1998
Recommend
Eddy
#include<stdio.h>
int s[500][4];
int main()
{
int P,i;
while (scanf("%d",&P)!=EOF)
{
if (P==0) return 0;
int N=(P-1)/4+1;
int M=4*N;
int l=1,r=M;
for (i=1;i<=N;i++)
{
s[i][0]=r--;
s[i][1]=l++;
s[i][2]=l++;
s[i][3]=r--;
}
printf("Printing order for %d pages:\n",P);
for (i=1;i<=N;i++)
{
if (!(s[i][0]>P && s[i][1]>P))
{
printf("Sheet %d, front:",i);
if (s[i][0]>P) printf(" Blank, ");
else printf(" %d, ",s[i][0]);
if (s[i][1]>P) printf("Blank\n");
else printf("%d\n",s[i][1]);
}
if (!(s[i][2]>P && s[i][3]>P))
{
printf("Sheet %d, back :",i);
if (s[i][2]>P) printf(" Blank, ");
else printf(" %d, ",s[i][2]);
if (s[i][3]>P) printf("Blank\n");
else printf("%d\n",s[i][3]);
}
}
}
return 0;
}
Booklet Printing[HDU1117]的更多相关文章
- ZOJ 1178 Booklet Printing
原题链接 题目大意:书本印刷都是用大开的纸张对折.比如一个册子一共4页,为了方便装订,外侧印刷1.4页,内侧印刷2.3页,这样对折之后就可以按照正常阅读习惯翻页了.此题目的就是给出书的总页数,要求计算 ...
- UVa 637 - Booklet Printing
题目:模拟输出n页书的装订打印状态. 分析:模拟.页数为(n+3)/ 4,仅仅有n不超过半篇时会输出半篇. 说明:好多曾经做过的题目(⊙_⊙). #include <cstdlib> #i ...
- POJ 题目分类(转载)
Log 2016-3-21 网上找的POJ分类,来源已经不清楚了.百度能百度到一大把.贴一份在博客上,鞭策自己刷题,不能偷懒!! 初期: 一.基本算法: (1)枚举. (poj1753,poj2965 ...
- HDU——PKU题目分类
HDU 模拟题, 枚举1002 1004 1013 1015 1017 1020 1022 1029 1031 1033 1034 1035 1036 1037 1039 1042 1047 1048 ...
- (转)POJ题目分类
初期:一.基本算法: (1)枚举. (poj1753,poj2965) (2)贪心(poj1328,poj2109,poj2586) (3)递归和分治法. (4)递推. ...
- poj分类
初期: 一.基本算法: (1)枚举. (poj1753,poj2965) (2)贪心(poj1328,poj2109,poj2586) (3)递归和分治法. ( ...
- 转载 ACM训练计划
leetcode代码 利用堆栈:http://oj.leetcode.com/problems/evaluate-reverse-polish-notation/http://oj.leetcode. ...
- poj 题目分类(1)
poj 题目分类 按照ac的代码长度分类(主要参考最短代码和自己写的代码) 短代码:0.01K--0.50K:中短代码:0.51K--1.00K:中等代码量:1.01K--2.00K:长代码:2.01 ...
- POJ题目分类(按初级\中级\高级等分类,有助于大家根据个人情况学习)
本文来自:http://www.cppblog.com/snowshine09/archive/2011/08/02/152272.spx 多版本的POJ分类 流传最广的一种分类: 初期: 一.基本算 ...
随机推荐
- HDU 1025 Constructing Roads In JGShining's Kingdom(求最长上升子序列nlogn算法)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1025 解题报告:先把输入按照r从小到大的顺序排个序,然后就转化成了求p的最长上升子序列问题了,当然按p ...
- MySQL数据库服务器的架设
导读 MySQL数据库是Linux操作系统上用得最多的数据库系统,它可以非常方便的与其它服务器集成在一起,如Apache.Vsftpd.Postfix等.下面介绍RHEL 6平台MySQL数据库服务器 ...
- ZeroMQ(java)之Router与Dealer运行原理
在开始这部分的内容之前,先来看看ZeroMQ中HWM概念---High-Water Marks 当系统的数据量很大,而且发送频率很高的情况下,内存就很重要了,如果处理不好会出现很多问题,例如如下场景: ...
- [ruby on rails] 跟我学之(2)HelloWorld
1. 创建工程 rails new blog 2.查看下文件结构 tree 输出如下,请留意红圈中的部分. Gemfile, 用来管理应用程序的gems, 有点类似于python的包,有专门的网站来查 ...
- HTML5/CSS3开发工具
1.谷歌Google Web Designer https://www.google.com/webdesigner/ 这个工具不能在xp上运行,可以在win7 win8上运行 http://www. ...
- 卸载Windows服务
在Windows中,有一类程序称为服务,在操作系统内核加载完成后就开始加载.这里程序往往运行在操作系统的底层,因此资源占用比较大.执行效率比较 高,比较有代表性的就是杀毒软件. 但是一旦因为特殊原因不 ...
- Java for LeetCode 043 Multiply Strings
Given two numbers represented as strings, return multiplication of the numbers as a string. Note: Th ...
- HDU 5793 A Boring Question (逆元+快速幂+费马小定理) ---2016杭电多校联合第六场
A Boring Question Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others ...
- 一、HTML和CSS基础--HTML+CSS基础课程--第5部分
第九章 CSS盒模型 元素分类 : 在讲解CSS布局之前,我们需要提前知道一些知识,在CSS中,html中的标签元素大体被分为三种不同的类型:块状元素.内联元素(又叫行内元素)和内联块状元素. 常用的 ...
- .net学习笔记----会话状态Session
一.会话状态Session Session用于服务器端状态管理,使用Session之后,每个客户端都可以将实际的数据保存在服务器上,对于每个客户端的数据,将会生成一个对应的唯一的key(保存在客户端) ...