2541: Paper Cutting 

Time Limit(Common/Java):1000MS/10000MS     Memory Limit:65536KByte
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

CTU Open 2003

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

  1. 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 ...

  2. 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) ...

  3. 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 ...

  4. 【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 ...

  5. AtCoder Beginner Contest 058 ABCD题

    A - ι⊥l Time limit : 2sec / Memory limit : 256MB Score : 100 points Problem Statement Three poles st ...

  6. TOJ 2944 Mussy Paper

    2944.   Mussy Paper Time Limit: 2.0 Seconds   Memory Limit: 65536K    Special JudgeTotal Runs: 381  ...

  7. CF Playing with Paper

    Playing with Paper time limit per test 2 seconds memory limit per test 256 megabytes input standard ...

  8. 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 ...

  9. 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

  1. IT人怎样防止过劳死?如何成为时间的主人?

    投行的朋友还没走几天,搜狐的一位同胞又去了.又是过劳死!    每当读到这类新闻,IT人无不反镜自照,顾影自怜.无法拼爹拼钱的我们,似乎只有拼命了.生活好惨淡啊!    有人说:年轻人,悠着点儿!立刻 ...

  2. Volley源码解析(三) 有缓存机制的情况走缓存请求的源码分析

    Volley源码解析(三) 有缓存机制的情况走缓存请求的源码分析 Volley之所以高效好用,一个在于请求重试策略,一个就在于请求结果缓存. 通过上一篇文章http://www.cnblogs.com ...

  3. 【读书笔记】构建之法(CH4~CH6)

    从chapter4至chapter6,围绕着构建过程的合作讨论构建之法,而合作与个人工作的区别却以一个微妙的问题为开端:阅读别人的代码有多难? 两人合作:(驾驶员与领航员) 合作要注意代码风格规范与设 ...

  4. 【R语言进行数据挖掘】决策树和随机森林

    1.使用包party建立决策树 这一节学习使用包party里面的函数ctree()为数据集iris建立一个决策树.属性Sepal.Length(萼片长度).Sepal.Width(萼片宽度).Peta ...

  5. HDU 1964 Pipes (插头DP,变形)

    题意:给一个n*m的矩阵,每个格子都是必走的,且无障碍格子,每对格子之间都有一个花费,问哈密顿回路的最小花费. 思路: 这个和Formula1差不多,只是求得是最小花费,这只需要修改一下DP值为花费就 ...

  6. iOS 查看包架构信息

    lipo -info libUMSocial_Sdk_4.2.a 查看包架构信息

  7. Python-OpenCV:cv2.imread(),cv2.imshow(),cv2.imwrite()

    为什么使用Python-OpenCV? 虽然python 很强大,而且也有自己的图像处理库PIL,但是相对于OpenCV 来讲,它还是弱小很多.跟很多开源软件一样OpenCV 也提供了完善的pytho ...

  8. java web.xml被文件加载过程及加载顺序小结

    web.xml加载过程(步骤): 1.启动WEB项目的时候,容器(如:Tomcat)会去读它的配置文件web.xml.读两个节点: <listener></listener> ...

  9. Workrave怎么用 Workrave使用方法, Workrave 健康计时器,预防电脑长期操作的职业病伤害

    下载绿色版: https://portableapps.com/apps/utilities/workrave_portable 选择阅读模式: 中文: 可以只选择启动休息的计时器,这样其他2个就不用 ...

  10. jquery动态实现填充下拉框

    当点下拉框时动态加载后台数据. 后台代码 protected void doPost(HttpServletRequest request, HttpServletResponse response) ...