北大poj-1005
I Think I Need a Houseboat
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 99271 | Accepted: 43245 |
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.
Source
#include <stdio.h> #define MAX_NUM 100 int gwNum;
float gwX[MAX_NUM];
float gwY[MAX_NUM];
int gwYear[MAX_NUM]; void GetData()
{
int i = ;
scanf("%d",&gwNum);
for(i=; i<gwNum; i++)
{
scanf("%f %f", &gwX[i], &gwY[i]);
}
} void Calc()
{
int i = ;
for(i=; i<gwNum; i++)
{
gwYear[i] = (int)(3.14*(gwX[i]*gwX[i]+gwY[i]*gwY[i])/) + ;
}
} void PrintResult()
{
int i = ;
for(i=; i<gwNum; i++)
{
printf("Property %d: This property will begin eroding in year %d.\n", i+, gwYear[i]);
}
printf("END OF OUTPUT.");
} int main(void)
{
GetData();
Calc();
PrintResult();
return ;
}
北大poj-1005的更多相关文章
- 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 ...
- 北大POJ题库使用指南
原文地址:北大POJ题库使用指南 北大ACM题分类主流算法: 1.搜索 //回溯 2.DP(动态规划)//记忆化搜索 3.贪心 4.图论 //最短路径.最小生成树.网络流 5.数论 //组合数学(排列 ...
- 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 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 ...
- [POJ 1005] I Think I Need a Houseboat C++解题
I Think I Need a Houseboat Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 81874 ...
- 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
简单题 #include <iostream> #include <stdio.h> #include <string.h> #include <stack& ...
- poj 1005 I Think I Need a Houseboat
#include <iostream> using namespace std; const double pi = 3.1415926535; int main() { ;; doubl ...
- [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 ...
随机推荐
- 分页组件 - layui.laypage
<!doctype html> <html> <head> <meta charset="utf-8"> <title> ...
- JQuery中操作Css样式的方法
JQuery中操作Css样式的方法//1.获取和设置样式 $("#tow").attr("class")获取ID为tow的class属性 $("#tw ...
- js中function函数
function:是具备某个功能的方法,方法本身没有意义,只有执行方法才有价值. function: 1 创建一个函数: 2 执行这个方法: 例: 创建 function 方法名(){ 存放某个功能的 ...
- html5对密码加密
今天找了几个关于对html5的密码加密的方法,仅供参考 1.base64加密:在页面中引入base64.js文件,调用方法为: <html> <head> <meta c ...
- Delphi与Windows 7下的用户账户控制(UAC)机制 及 禁用兼容性助手
WIN7, Vista提供的UAC机制,它的主要目的是防止对于操作系统本身的恶意修改.对于Delphi程序的影响,UAC主要在于以下几点:1.由于UAC机制,Delphi对于系统的操作可能无声的失败, ...
- autoloader
自动加载 $loader = new Zend_Application_Module_Autoloader(array( 'namespace' => 'Blog', 'base ...
- zend studio 9 字体,颜色,快捷键等相关设置
1.zend studio 9可以破解吗? 可以的,具体破解步骤查看:http://www.geekso.com/ZendStudio9-key/ 2.如何将zend studio 9的默认GBK编码 ...
- Linux内核原子(1) - spinlock的实现
spinlock的数据结构spinlock_t定义在头文件linux/spinlock_types.h里面: typedef struct { raw_spinlock_t raw_lock; #if ...
- Powerdesigner SqlServer转Oracle(转)
参考文章,原文地址:http://blog.csdn.net/cicada688/article/details/7802881 问题1:sqlserver数据库直接转oracle.字段类型由sql ...
- kaggle数据挖掘竞赛初步--Titanic<原始数据分析&缺失值处理>
Titanic是kaggle上的一道just for fun的题,没有奖金,但是数据整洁,拿来练手最好不过啦. 这道题给的数据是泰坦尼克号上的乘客的信息,预测乘客是否幸存.这是个二元分类的机器学习问题 ...