[POJ 1005] I Think I Need a Houseboat C++解题
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 81874 | Accepted: 35368 |
Description
After doing more research, Fred has learned that the land that is
being lost forms a semicircle. This semicircle is part of a circle
centered at (0,0), with the line that bisects the circle being the X
axis. Locations below the X axis are in the water. The semicircle has an
area of 0 at the beginning of year 1. (Semicircle illustrated in the
Figure.)
Input
first line of input will be a positive integer indicating how many data
sets will be included (N). Each of the next N lines will contain the X
and Y Cartesian coordinates of the land Fred is considering. These will
be floating point numbers measured in miles. The Y coordinate will be
non-negative. (0,0) will not be given.
Output
each data set, a single line of output should appear. This line should
take the form of: “Property N: This property will begin eroding in year
Z.” Where N is the data set (counting from 1), and Z is the first year
(start from 1) this property will be within the semicircle AT THE END OF
YEAR Z. Z must be an integer. After the last data set, this should
print out “END OF OUTPUT.”
Sample Input
2
1.0 1.0
25.0 0.0
Sample Output
Property 1: This property will begin eroding in year 1.
Property 2: This property will begin eroding in year 20.
END OF OUTPUT.
Hint
2.This problem will be judged automatically. Your answer must match
exactly, including the capitalization, punctuation, and white-space.
This includes the periods at the ends of the lines.
3.All locations are given in miles.
解决思路
这是一道很简单的计算几何题,本质就是计算点是否在指定的半圆内。
源码
/*
poj 1000
version:1.0
author:Knight
Email:S.Knight.Work@gmail.com
*/ #include<cstdio>
#include<cmath>
using namespace std;
const double PI = 3.1415926;
int main(void)
{
double S;//被腐蚀陆地的总面积
double r;//腐蚀陆地的半径
int i,j;
int T;
double X,Y;//Fred的坐标
double Distance;//Fred据坐标原点的距离
scanf("%d", &T);
for (i=; i<=T; i++)
{
scanf("%lf%lf", &X, &Y);
Distance = sqrt(X * X + Y * Y);
S = 0.0;
for (j=; ; j++)
{
S += 50.0;
r = sqrt( * S / PI);
if (r >= Distance)
{
break;
}
}
printf("Property %d: This property will begin eroding in year %d.\n", i, j);
}
printf("END OF OUTPUT.\n");
return ;
}
[POJ 1005] I Think I Need a Houseboat C++解题的更多相关文章
- POJ. 1005 I Think I Need a Houseboat(水 )
POJ. 1005 I Think I Need a Houseboat(水 ) 代码总览 #include <cstdio> #include <cstring> #incl ...
- OpenJudge/Poj 1005 I Think I Need a Houseboat
1.链接地址: http://bailian.openjudge.cn/practice/1005/ http://poj.org/problem?id=1005 2.题目: I Think I Ne ...
- poj 1005:I Think I Need a Houseboat(水题,模拟)
I Think I Need a Houseboat Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 85149 Acce ...
- [POJ] #1005# I Think I Need a Houseboat : 浮点数运算
一. 题目 I Think I Need a Houseboat Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 97512 ...
- poj 1005 I Think I Need a Houseboat
#include <iostream> using namespace std; const double pi = 3.1415926535; int main() { ;; doubl ...
- I Think I Need a Houseboat POJ - 1005
I Think I Need a Houseboat POJ - 1005 解题思路:水题 #include <iostream> #include <cstdio> #inc ...
- I Think I Need a Houseboat POJ - 1005(数学)
题目大意 在二维坐标内选定一个点,问你当洪水以半圆形扩散且每年扩散50单位,哪一年这个点被被洪水侵蚀? 解法 代码 #include <iostream> #include <cst ...
- POJ 1005 解题报告
1.题目描述 2.解题思路 好吧,这是个水题,我的目的暂时是把poj第一页刷之,所以水题也写写吧,这个题简单数学常识而已,给定坐标(x,y),易知当圆心为(0,0)时,半圆面积为0.5*PI*(x ...
- 1005 -- I Think I Need a Houseboat
I Think I Need a Houseboat Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 105186 Acc ...
随机推荐
- ThreadLocal(关于struts2的ThreadLocal,实际上Jdk1.2就有了)
ThreadLocal是通过在不同线程中操作变量的副本,来达到线程安全的目的,是用空间资源换时间资源的方式.今天在看struts2源码的时候,发现ActionContext中,就持有一个静态的Thre ...
- DetachedCriteria的简单使用
一. DetachedCriteria使得hibernate能够对查询条件进行面向对象的方式来组装.其创建方式有两种: 1.1直接用class创建:DetachedCriteria criteria ...
- JavaScript中登录名的正则表达式及解析(0基础)
简言 在JavaScript中,经常会用到正则表达式来进行模式匹配.例如,登录名验证,密码强度验证,字符串查找或替换等操作.现在就开始吧,零基础写出你的第一个正则表达式! 在做用户注册时,都会用到登录 ...
- Android ORM对象关系映射之GreenDAO建立多表关联
https://blog.csdn.net/u010687392/article/details/48496299 利用GreenDAO可以非常方便的建立多张表之间的关联 一对一关联 通常我们在操作数 ...
- Ionic 2 中的创建一个闪视卡片组件
闪视卡片是记忆信息的重要工具,它的使用可以追溯到19世纪.我们将要创建一个很酷的短暂动画来实现它.看起来像是这个样子的: 闪视卡片示例 Ionic 2 实例开发 新增章节将为你介绍如何在Ionic 2 ...
- HDU 2188 悼念512汶川大地震遇难同胞——选拔志愿者(巴什博弈)
思路:若能给对方留下m+1,就可以胜.否则败. #include <iostream> using namespace std; int main() { int t,n,m;cin> ...
- uva297 Quadtrees (线段树思想,区间操作)
借鉴了线段数区间操作的思想,只是把一个结点的孩子扩展到了4个, 结点k,四个孩子编号分别为4*k+1,4*k+2,4*k+3,4*K+4,从零开始. 根据层数,确定权值. #include<cs ...
- python_86_shutil模块
#高级的文件.文件夹.压缩包.处理模块 import shutil f1=open('sys模块.py','r',encoding='utf-8') f2=open('copy1.py','w',en ...
- Golang glog使用详解
golang/glog 是 C++ 版本 google/glog 的 Go 版本实现,基本实现了原生 glog 的日志格式.在 Kuberntes 中,glog 是默认日志库. glog 的使用与特性 ...
- ReactiveCocoa入门-part2
ReactiveCocoa是一个框架,它能让你在iOS应用中使用函数响应式编程(FRP)技术.在本系列教程的第一部分中,你学到了如何将标准的动作与事件处理逻辑替换为发送事件流的信号.你还学到了如何转换 ...