Sdut 2416 Fruit Ninja II(山东省第三届ACM省赛 J 题)(解析几何)
Time Limit: 5000MS Memory limit: 65536K
题目描述
Haveyou ever played a popular game named "Fruit Ninja"?
Fruit Ninja (known as Fruit Ninja HD on the iPad and Fruit Ninja THD for NvidiaTegra 2 based Android devices) is a video game developed by Halfbrick. It wasreleased April 21, 2010 for iPod Touch and iPhone devices, July 12, 2010 forthe iPad, September 17, 2010
for Android OS devices. Fruit Ninja was wellreceived by critics and consumers. The iOS version sold over 200,000 copies inits first month. By March 2011 total downloads across all platforms exceeded 20million. It was also named one of Time magazine's 50 Best
iPhone Apps of 2011.
"Swipeyour finger across the screen to deliciously slash and splatter fruit like atrue ninja warrior. Be careful of bombs - they are explosive to touch and willput a swift end to your juicy adventure!" - As it described
onhttp://www.fruitninja.com/, in Fruit Ninja the player slices fruit with a bladecontrolled via a touch pad. As the fruit is thrown onto the screen, the player swipestheir finger across the screen to create a slicing motion, attempting to slicethe fruit in
parts. Extra points are awarded for slicing multiple fruits withone swipe, and players can use additional fingers to make multiple slicessimultaneously. Players must slice all fruit; if three pieces of fruit aremissed the game ends. Bombs are occasionally
thrown onto the screen, and willalso end the game should the player slice them.
Maybe you are an excellent player of Fruit Ninja, but in this problem we focus onsomething more mathematically. Consider a certain slicing trace you create onthe touch pad, you slice a fruit (an apple or a banana or something else) intotwo parts at once. Can
you figure out the volume of each part?
Impossibletask? Let us do some simplification by define our own Fruit Ninja game.
In our new Fruit Ninja game, only one kind of fruit will be thrown into the air- watermelon. What's more, the shape of every watermelon is a special Ellipsoid(details reachable at http://en.wikipedia.org/wiki/Ellipsoid) that it's polarradius OC is always equals
to it's equatorial radius OB. Formally, we can getthis kind of solid by revolving a certain ellipse on the x-axis. And theslicing trace the player created (represented as MN in Illustration III) is aline parallel to the x-axis. The slicing motion slice the
watermelon into twoparts, and the section (shown as the dark part in Illustration III) is parallelto plane x-O-y.
Given the length of OA, OB, OM (OM is thedistance between the section and plane x-O-y), your task is to figure out thevolume of the bigger part.
输入
Thereare multiple test cases. First line is an integer T (T ≈ 100), indicating thenumber of test cases.
For each test case, there are three integers: a, b, H, corresponding the lengthof OA, OB, OM. You may suppose that0 < b <= a <= 100 and 0 <= H <= 100.
输出
Outputcase number "Case %d: " followed by a floating point number (round to3) for each test case.
示例输入
4
2 2 0
2 2 1
2 2 2
2 2 3
示例输出
Case 1: 16.755
Case 2: 28.274
Case 3: 33.510
Case 4: 33.510
时间紧急,图片处理的不是太好,大家将就看吧
/***********************
刚看这个题目确实被吓到了,仔细看了以后你会发现这就是一个高数上的三重积分的问题。
椭球长半轴 a ,短半轴 b ,高 h,(h 就是 题中 的 h)
则椭球缺体积可以这样求:(Word 里的数学符号不能显示,截个图片吧)
PI 为圆周率
然后用这个体积加上半球的体积就OK了。。
公式: V = 2/3*PI*a*b*b+PI *a*(b*h—h^3/(3*b))
Code:
#include <iostream>
#include<stdio.h>
using namespace std;
const double PI = 3.14159265358; //PI = 3.1415926535 时 WA,所以注意精度
int main()
{
int t,count = 1;;
double a,b,h,V;
scanf("%d",&t);
while(t--)
{
scanf("%lf%lf%lf",&a,&b,&h);
if(h>=b)
h = b;
V = 2.0/3.0*PI*a*b*b+PI*a*b*h-PI*a*h*h*h/(3.0*b);
printf("Case %d: ",count++);
printf("%.3lf\n",V);
}
return 0;
}
Sdut 2416 Fruit Ninja II(山东省第三届ACM省赛 J 题)(解析几何)的更多相关文章
- sdut 2416:Fruit Ninja II(第三届山东省省赛原题,数学题)
Fruit Ninja II Time Limit: 5000MS Memory limit: 65536K 题目描述 Have you ever played a popular game name ...
- SDUT 2416:Fruit Ninja II
Fruit Ninja II Time Limit: 5000MS Memory limit: 65536K 题目描述 Have you ever played a popular game name ...
- Sdut 2409 The Best Seat in ACM Contest(山东省第三届ACM省赛 H 题)(模拟)
题目描述 Cainiao is a university student who loves ACM contest very much. It is a festival for him once ...
- 山东省第三届ACM省赛
Solved ID PID Title Accepted Submit A 2407 Impasse (+) 0 0 B 2415 Chess 0 0 C 2414 An interest ...
- sdut2408 pick apples (贪心+背包)山东省第三届ACM省赛
版权声明:本文为博主原创文章.未经博主同意不得转载. https://blog.csdn.net/svitter/article/details/24642587 本文出自:http://blog.c ...
- 青岛理工ACM交流赛 J题 数格子算面积
数格子算面积 Time Limit: 1000MS Memory limit: 262144K 题目描述 给你一个多边形(用’\’和’/’表示多边形的边),求多边形的面积. 输入 第一行两个正整数h ...
- 福建工程学院第十四届ACM校赛J题题解
第六集,想不到你这个浓眉大眼的都叛变革命了 题意: 给你两个只包含01的字符串S和T,问你在允许一次错误的情况下,T是否能成为S的子串 思路: 这个问题的解法挺多,我是用fft匹配的,也比较简单,针对 ...
- [原]sdut2624 Contest Print Server (大水+大坑)山东省第四届ACM省赛
本文出自:http://blog.csdn.net/svitter 原题:http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&am ...
- [2012山东省第三届ACM大学生程序设计竞赛]——n a^o7 !
n a^o7 ! 题目:http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2413 Time Lim ...
随机推荐
- Python实现二叉树的前序遍历、中序遍历
计算根节点到叶子节点的所组成的数字(1247, 125, 1367)以及叶子节点到根节点组成的数字(7421, 521, 8631),其二叉树树型结构如下 计算从根节点到叶子节点组成的数字,本质上来说 ...
- spring的有状态BEAN和无状态BEAN
有状态会话bean :每个用户有自己特有的一个实例,在用户的生存期内,bean保持了用户的信息,即“有状态”:一旦用户灭亡(调用结束或实例结束),bean的生命期也告结束.即每个用户最初都会得到一 ...
- GNU LIBC源代码学习之strcmp
比較两个字符串 我的代码块 #include <string.h> int my_strcmp(const char* s1,const char * s2) { if((s1==NULL ...
- Android UI开发第三十三篇——Navigation Drawer For Android API 7
Creating a Navigation Drawer中使用的Navigation Drawer的android:minSdkVersion="14",现在Android API ...
- 【Android UI设计与开发】之具体解释ActionBar的使用
具体解释Android中的ActionBar的使用 请尊重他人的劳动成果,转载请注明出处:具体解释Android中的ActionBar的使用 http://blog.csdn.net/fengyuzh ...
- android颜色对应的xml配置值,颜色表
网上找的一些颜色值 XML配置 <?xml version="1.0" encoding="utf-8" ?> <resources> ...
- count(1) count(*)
mysql from t; +---+ | +---+ | | | | +---+ rows in set (0.00 sec) mysql) from t; +----------+ ) | +-- ...
- ASP.NET 未被授权访问所请求的资源。请考虑授予 ASP.NET 请求标识访问此资源的权限
开发了一个导入TXT文件的功能,执行过程中出错.提示:.....ASP.NET 未被授权访问所请求的资源.请考虑授予 ASP.NET 请求标识访问此资源的权限.ASP.NET 有一个在应用程序没有模拟 ...
- 关于NSRunLoop和NSTimer的深入理解
一.什么是NSRunLoop NSRunLoop是消息机制的处理模式 NSRunLoop的作用在于有事情做的时候使的当前NSRunLoop的线程工作,没有事情做让当前NSRunLoop的线程休眠 NS ...
- Android5.0之CardView的使用
CardView也是一个非常炫酷的控件,一般我们将CardView配合RecyclerView来使用,当然,CardView也可以配合ListView来使用,都是可以的.OK,我们先来看一张CardV ...