1012. Stacking Cylinders

最新推荐文章于 2019-10-23 23:03:59 发布

原创最新推荐文章于 2019-10-23 23:03:59 发布 · 265 阅读

0 ·

CC 4.0 BY-SA版权

sicily 专栏收录该内容

13 篇文章

订阅专栏

本文介绍了一个程序设计问题，即如何通过已知底部圆柱的坐标来计算顶部圆柱中心的坐标。该问题涉及数学计算，特别是利用几何公式进行逐级累加求解。输入包括底部圆柱的数量及它们的横坐标，输出则是顶部圆柱的精确坐标。

1012. Stacking Cylinders

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

Cylinders (e.g. oil drums) (of radius 1 foot) are stacked in a rectangular bin. Each cylinder on an upper row rests on two cylinders in the row below. The cylinders in the bottom row rest on the floor. Each row has one less cylinder than the row below.

This problem is to write a program to compute the location of the center of the top cylinder from the centers of the cylinders on the bottom row. Computations of intermediate values should use double precision.

Input

Each data set will appear in one line of the input. An input line consists of the number, n, of cylinders on the bottom row followed by n floating point values giving the x coordinates of the centers of the cylinders (the y coordinates are all 1.0 since the cylinders are resting on the floor (y = 0.0)). The value of n will be between 1 and 10 (inclusive). The end of input is signaled by a value of n = 0. The distance between adjacent centers will be at least 2.0 (so the cylinders do not overlap) but no more than 3.4 (cylinders at level k will never touch cylinders at level k – 2).

Output

The output for each data set is a line containing the x coordinate of the topmost cylinder rounded to 4 decimal places, a space and the y coordinate of the topmost cylinder to 4 decimal places. Note: To help you check your work, the x-coordinate of the center of the top cylinder should be the average of the x-coordinates of the leftmost and rightmost bottom cylinders.

Sample Input

4 1.0 4.4 7.8 11.21 1.06 1.0 3.0 5.0 7.0 9.0 11.010 1.0 3.0 5.0 7.0 9.0 11.0 13.0 15.0 17.0 20.45 1.0 4.4 7.8 14.6 11.20

Sample Output

6.1000 4.16071.0000 1.00006.0000 9.660310.7000 15.91007.8000 5.2143

// 利用公式 y += sqrt(4 - (a[i + 1] - a[i]) * (a[i + 1] - a[i])/4) 加上去就好了

#include <iostream>
#include <algorithm>
#include <iomanip>
#include <cmath>

using namespace std;

int main()
{
	int n;
	while (cin >> n && n != 0)
	{
		double a[1000];
		double x = 0.0, y = 1.0;
		for (int i = 0; i < n; i++)
			cin >> a[i];
		sort(a, a + n);
		x = (a[0] + a[n - 1])/2;
		for (int i = 0; i < n - 1; i++)
		{
			y += sqrt(4 - (a[i + 1] - a[i]) * (a[i + 1] - a[i])/4);
		}
		cout << fixed << setprecision(4) << x << " ";
		cout << fixed << setprecision(4) << y << endl;
	}

	return 0;
}