http://acm.hdu.edu.cn/status.php

Alfredo's Pizza Restaurant

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1732    Accepted Submission(s): 1023

Problem Description
Traditionally after the Local Contest, judges and contestants go to their favourite restaurant, Alfredos Pizza Restaurant. The contestants are really hungry after trying hard for five hours. To get their pizza as quickly as possible, they just decided to order one big pizza for all instead of several small ones. They wonder whether it is possible to put the big rectangular pizza on the surface of the round table such that it does not overhang the border of the table. Write a program that helps them!
 
Input
The input file contains several test cases. Each test case starts with an integer number r, the radius of the surface of the round table the contestants are sitting at. Input is terminated by r=0. Otherwise, 1 ≤ r ≤ 1000. Then follow 2 integer numbers w and l specifying the width and the length of the pizza, 1 ≤ w ≤ l ≤ 1000.
 
Output
Output for each test case whether the ordered pizza will fit on the table or not. Adhere to the format shown in the sample output. A pizza which just touches the border of the table without intersecting it is considered fitting on the table, see example 3 for clarification.
 
Sample Input
38 40 60 
 35 20 70
50 60 80 0
 
Sample Output
Pizza 1 fits on the table.
Pizza 2 does not fit on the table.
Pizza 3 fits on the table.
 #include<stdio.h>
int main()
{
int t=;
double r,w,l;
while(scanf("%lf",&r)&&r!=)
{
scanf("%lf%lf",&w,&l);
w=w/;
l=l/;
if(w*w+l*l<=r*r)
printf("Pizza %d fits on the table.\n",t++);
else
printf("Pizza %d does not fit on the table.\n",t++);
}
return ;
}
 

HDU-2368 Alfredo's Pizza Restaurant的更多相关文章

  1. hdu2368Alfredo's Pizza Restaurant

    Problem Description Traditionally after the Local Contest, judges and contestants go to their favour ...

  2. 转载:hdu 题目分类 (侵删)

    转载:from http://blog.csdn.net/qq_28236309/article/details/47818349 基础题:1000.1001.1004.1005.1008.1012. ...

  3. 杭电ACM分类

    杭电ACM分类: 1001 整数求和 水题1002 C语言实验题——两个数比较 水题1003 1.2.3.4.5... 简单题1004 渊子赛马 排序+贪心的方法归并1005 Hero In Maze ...

  4. TZOJ 2289 Help Bob(状压DP)

    描述 Bob loves Pizza but is always out of money. One day he reads in the newspapers that his favorite ...

  5. HDU 4883 TIANKENG’s restaurant

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4883 解题报告:一家餐馆一天中有n波客人来吃饭,第 i 波  k 客人到达的时间是 s ,离开时的时间 ...

  6. HDU 1103 Flo's Restaurant(模拟+优先队列)

    Flo's Restaurant Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) ...

  7. TIANKENG’s restaurant HDU - 4883 (暴力)

    TIANKENG manages a restaurant after graduating from ZCMU, and tens of thousands of customers come to ...

  8. HDU 4883 TIANKENG’s restaurant Bestcoder 2-1(模拟)

    TIANKENG's restaurant Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/O ...

  9. HDU 4886 TIANKENG’s restaurant(Ⅱ) ( 暴力+hash )

    TIANKENG’s restaurant(Ⅱ) Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 130107/65536 K (Ja ...

随机推荐

  1. mongodb下载及安装配置教程【仅供参考】

    1 下载 下载页面地址:https://www.mongodb.org/downloads 版本选择:电脑系统是64位的,所以我选择了 Windows 64-bit 2008 R2+ ,msi包 2 ...

  2. 【BZOJ3884】【降幂大法】上帝与集合的正确用法

    Description 根据一些书上的记载,上帝的一次失败的创世经历是这样的: 第一天, 上帝创造了一个世界的基本元素,称做“元”. 第二天, 上帝创造了一个新的元素,称作“α”.“α”被定义为“元” ...

  3. Sdut 2164 Binomial Coeffcients (组合数学) (山东省ACM第二届省赛 D 题)

    Binomial Coeffcients TimeLimit: 1000ms   Memory limit: 65536K  有疑问?点这里^_^ 题目描述 输入 输出 示例输入 1 1 10 2 9 ...

  4. gcc命令以及makefile文件

    (一)makefile里涉及到的gcc命令 gcc -I./inc:指定头文件寻找目录 将按照 ./inc --> /usr/include --> /usr/local/include的 ...

  5. 如何判断一个Div是否在可视区域,判断div是否可见

    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/ ...

  6. 小笔记(三):PHP使用thinkphp3.2.3对数组进行分页

    之前写过thinkphp3.2.3直接在查询数据的时候进行分页,前段时间用到了将查询之后的数组进行整理后进行分页,用到的一个函数array_slice($arr, $start, $length,tr ...

  7. VS2015安装开发ios android

    前几天很火,装了一下,结果是不是太满意,装了VS2015只是多了一个android和ios的模版,最终还是要装xamarin ,最后装了个xamarin ,然后破解 破解地址:http://www.c ...

  8. POJ 1860 Currency Exchange 毫无优化的bellman_ford跑了16Ms,spfa老是WA。。

    题目链接: http://poj.org/problem?id=1860 找正环,找最长路,水题,WA了两天了.. #include <stdio.h> #include <stri ...

  9. 单片微机原理P0:80C51结构原理

    本来我真的不想让51的东西出现在我的博客上的,因为51这种东西真的太low了,学了最多就所谓的垃圾科创利用一下,但是想一下这门课我也要考试,还是写一点东西顺便放博客上吧. 这一系列主要参考<单片 ...

  10. 时序图(Sequence Diagram)

    控制焦点Focus on Control 的取值: Alternative fragment(denoted “alt”) 与 if…then…else对应 Option fragment (deno ...