[POJ] 1005 I Think I Need a Houseboat

最新推荐文章于 2023-08-04 09:51:04 发布

RoderickLi

最新推荐文章于 2023-08-04 09:51:04 发布

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分类专栏： OJ 文章标签： POJ

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本文介绍了一个简单的算法，用于预测在Louisiana州由于密西西比河侵蚀造成的土地损失对特定房产的影响。该算法通过计算输入坐标的半圆半径和面积，确定房产开始遭受侵蚀的年份。

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I Think I Need a Houseboat

Time Limit: 1000MS	Memory Limit: 10000K
Total Submissions: 115914	Accepted: 49974

Description

Fred Mapper is considering purchasing some land in Louisiana to build his house on. In the process of investigating the land, he learned that the state of Louisiana is actually shrinking by 50 square miles each year, due to erosion caused by the Mississippi River. Since Fred is hoping to live in this house the rest of his life, he needs to know if his land is going to be lost to erosion.

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

The 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

For 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.

解题思路

简单题。通过计算输入坐标点所在半圆的半径，进而计算出半圆面积，int(半圆面积/50+1)即为所求。

#include <iostream>
#define PI 3.1415926
using namespace std;

int main()
{
    int n;
    cin>>n;
    for(int i=1;i<=n;i++){
        float x,y,s;
        cin>>x>>y;
        s=(x*x+y*y)*PI/2;
        cout<<"Property "<<i<<": This property will begin eroding in year "<<(int)(s/50+1)<<"."<<endl;
    }
    cout<<"END OF OUTPUT."<<endl;
    return 0;
}