TOJ 2541: Paper Cutting
2541: Paper Cutting
Total Submit: 1 Accepted:1
Description
ACM managers need business cards to present themselves to their customers and partners. After the cards are printed on a large sheet of paper, they are cut with a special cutting machine. Since the machine operation is very expensive, it is necessary to minimize the number of cuts made. Your task is to find the optimal solution to produce the business cards.
There are several limitations you have to comply with. The cards are always printed in a grid structure of exactly a * b cards. The structure size (number of business cards in a single row and column) is fixed and cannot be changed due to a printing software restrictions. The sheet is always rectangular and its size is fixed. The grid must be perpendicular to the sheet edges, i.e., it can be rotated by 90 degrees only. However, you can exchange the meaning of rows and columns and place the cards into any position on the sheet, they can even touch the paper edges.
For instance, assume the card size is 3 * 4 cm, and the grid size 1 * 2 cards. The four possible orientations of the grid are depicted in the following figure. The minimum paper size needed for each of them is stated.
The cutting machine used to cut the cards is able to make an arbitrary long continuous cut. The cut must run through the whole piece of the paper, it cannot stop in the middle. Only one free piece of paper can be cut at once -- you cannot stack pieces of paper onto each other, nor place them beside each other to save cuts.
Input
The input consists of several test cases. Each of them is specified by six positive integer numbers, A,B,C,D,E,F, on one line separated by a space. The numbers are:
A and B are the size of a rectangular grid, 1 <= A,B <= 1 000,
C and D are the dimensions of a card in cms, 1 <= C,D <= 1 000, and
E and F are the dimensions of a paper sheet in cms, 1 <= E,F <= 1 000 000.
The input is terminated by a line containing six zeros.
Output
For each of the test cases, output a single line. The line should contain the text: "The minimum number of cuts is X.", where X is the minimal number of cuts required. If it is not possible to fit the card grid onto the sheet, output the sentence "The paper is too small." instead.
Sample Input
Sample Output
Source
Tag
要切一种卡片 纸张大小事px,py
格子大小gx,gy 卡片要cx,xy;
于是每次就要切出gx*gy个大小为cx*cy的卡片 问最少切几次
关键是:
1:一次只能切连续,不能在中间中断 故只能一横或一竖切下来
2:每次只能切连在一起的纸 若你之前被切开的则不能一起切 当然纸张更不能折叠来切
(观察sample就可以知道了,然后对于大小为n*m的格子只要切n*m-1次就行了当然还要再检验一下边角需不需要切)
枚举所有情况啊
#include<stdio.h>
int a,b,c,d,e,f;
const int INF=<<;
int la(int w,int x,int y,int z)
{
if(w*y>e||x*z>f)return INF;
return w*x-+(w*y<e)+(x*z<f);
}
int main()
{
while(~scanf("%d%d%d%d%d%d",&a,&b,&c,&d,&e,&f),a||b||c||d||e||f)
{
int m=la(a,b,c,d),x;
if((x=la(b,a,c,d))<m)m=x;
if((x=la(a,b,d,c))<m)m=x;
if((x=la(b,a,d,c))<m)m=x;
if(m==INF)printf("The paper is too small.\n");
else printf("The minimum number of cuts is %d.\n",m);
}
return ;
}
TOJ 2541: Paper Cutting的更多相关文章
- HOJ题目分类
各种杂题,水题,模拟,包括简单数论. 1001 A+B 1002 A+B+C 1009 Fat Cat 1010 The Angle 1011 Unix ls 1012 Decoding Task 1 ...
- Chinese culture
文房四宝 笔墨纸砚是中国古代文人书房中必备的宝贝,被称为“文房四宝”.用笔墨书写绘画在 中国可追溯到五千年前.秦(前221---前206)时已用不同硬度的毛和竹管制笔:汉代(前206—公元220) ...
- OJ题解记录计划
容错声明: ①题目选自https://acm.ecnu.edu.cn/,不再检查题目删改情况 ②所有代码仅代表个人AC提交,不保证解法无误 E0001 A+B Problem First AC: 2 ...
- 【AtCoder】KEYENCE Programming Contest 2019
A - Beginning 这个年份恐怕需要+2 #include <bits/stdc++.h> #define fi first #define se second #define p ...
- AtCoder Beginner Contest 058 ABCD题
A - ι⊥l Time limit : 2sec / Memory limit : 256MB Score : 100 points Problem Statement Three poles st ...
- TOJ 2944 Mussy Paper
2944. Mussy Paper Time Limit: 2.0 Seconds Memory Limit: 65536K Special JudgeTotal Runs: 381 ...
- CF Playing with Paper
Playing with Paper time limit per test 2 seconds memory limit per test 256 megabytes input standard ...
- Codeforces Round #296 (Div. 2) A. Playing with Paper
A. Playing with Paper One day Vasya was sitting on a not so interesting Maths lesson and making an o ...
- Cutting Codeforces Round #493 (Div. 2)
Cutting There are a lot of things which could be cut — trees, paper, “the rope”. In this problem you ...
随机推荐
- @RequestParam和@ResponseBody注解的区别(转)
@RequestParam 用来处理Content-Type: 为 application/x-www-form-urlencoded编码的内容.(Http协议中,如果不指定Content-Type, ...
- .NET Core 1.0 CentOS7 尝试(二、VSCode安装)
晚上继续摸索~~ 一.VSCode安装 下载VSCode,地址:https://code.visualstudio.com/Docs/?dv=linux64_rpm 直接双击安装或者使用rpm -i ...
- 【CSS】纯css实现立体摆放图片效果
1. 元素的 width/height/padding/margin 的百分比基准 设置 一个元素 width/height/padding/margin 的百分比的时候,大家可知道基准是什么? 举 ...
- Android笔记--Bitmap
Android | Bitmap解析 Android中Bitmap是对图像的一种抽象.通过他可以对相应的图像进行剪裁,旋转,压缩,缩放等操作.这里循序渐进的一步步了解Bitmap的相关内容. 先了解B ...
- (转)SpringMVC学习(三)——SpringMVC的配置文件
http://blog.csdn.net/yerenyuan_pku/article/details/72231527 读者阅读过SpringMVC学习(一)——SpringMVC介绍与入门这篇文章后 ...
- [洛谷P4556][BZOJ3307]雨天的尾巴-T3订正
线段树合并+树上差分 题目链接(···) 「简单」「一般」——其实「一般」也只多一个离散化 考试时想法看>这里< 总思路:先存所有的操作,离散化,然后用树上差分解决修改,用权值线段树维护每 ...
- myna代码
https://github.com/TalkingData/Myna/tree/master/Dataset https://github.com/TalkingData/Myna
- UEditor1.4.3的实例程序
官网:http://ueditor.baidu.com/website/ 配置下就可以使用 (1)下载,解压后文件结构如下: (2)将整个文件夹改名ueditor后复制到WebRoot目录下: (3) ...
- common-fileupload上传图片并显示图片
效果图如下: 代码: 注意:需要jar包:commons-fileupload-1.2.1.jar 和 commons-io-1. ...
- spring源码学习之容器的基本实现
最近想拿出一部分时间来学习一下spring的源码,还特意买了一本书结合来看,当然主要是学习并跟着作者的思路来踏上学习spring的源码的道路,特意在此记录一下,<spring源码深度解析> ...