找出对角线倾斜且排成一排的正方形的一边
这里给出的是 n 个正方形，它们是倾斜的，在顶点处相互接触，并排成一排。给出了第一个正方形和最后一个正方形的中心之间的距离。正方形的边长相等。任务是找到每个正方形的边。 例:
输入: d = 42，n = 4 输出:每个方块的边长为 9.899 输入: d = 54，n = 7 输出:每个方块的边长为 6.364
接近 : 有 n 个正方形，每个正方形的边长为 a，第一个正方形和最后一个正方形之间的距离等于 d。从图中可以清楚地看出，它们是由对角线连接的。每条对角线的长度等于一√2 。 对于第一个和最后一个正方形，长度 d 下只覆盖对角线的一半，对于其余(n-2)个正方形，完整的对角线被 d 覆盖，因此 a 和 d 的关系如下:
a/√2+a/√2+(n-2) a√2 = d =>a√2+√2na–2a√2 = d =>n√2a–a√2 = d =>a = d/((n-1)(√2)) 正方形的边=中心之间的距离/((正方形的数目-1) * sqrt(2))。
以下是上述方法的实现:
C++
// C++ program to find side of the squares
// inclined and touch each other externally
// at vertices and are lined in a row
// and distance between the
// centers of first and last squares is given
#include <bits/stdc++.h>
using namespace std;
void radius(double n, double d)
{
cout << "The side of each square is "
<< d / ((n - 1) * sqrt(2)) << endl;
}
// Driver code
int main()
{
double d = 42, n = 4;
radius(n, d);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find side of the squares
// inclined and touch each other externally
// at vertices and are lined in a row
// and distance between the
// centers of first and last squares is given
import java.io.*;
class GFG
{
static void radius(double n, double d)
{
System.out.print( "The side of each square is "+
d / ((n - 1) * Math.sqrt(2)));
}
// Driver code
public static void main (String[] args)
{
double d = 42, n = 4;
radius(n, d);
}
}
// This code is contributed by anuj_67..
Python 3
# Python program to find side of the squares
# inclined and touch each other externally
# at vertices and are lined in a row
# and distance between the
# centers of first and last squares is given
def radius(n, d):
print("The side of each square is ",
d / ((n - 1) * (2**(1/2))));
# Driver code
d = 42; n = 4;
radius(n, d);
# This code is contributed by Rajput-Ji
C
// C# program to find side of the squares
// inclined and touch each other externally
// at vertices and are lined in a row
// and distance between the
// centers of first and last squares is given
using System;
class GFG
{
static void radius(double n, double d)
{
Console.WriteLine( "The side of each square is "+
d / ((n - 1) * Math.Sqrt(2)));
}
// Driver code
public static void Main ()
{
double d = 42, n = 4;
radius(n, d);
}
}
// This code is contributed by anuj_67..
java 描述语言
<script>
// javascript program to find side of the squares
// inclined and touch each other externally
// at vertices and are lined in a row
// and distance between the
// centers of first and last squares is given
function radius(n , d)
{
document.write( "The side of each square is "+
(d / ((n - 1) * Math.sqrt(2))).toFixed(5));
}
// Driver code
var d = 42, n = 4;
radius(n, d);
// This code is contributed by Amit Katiyar
</script>
Output:
The side of each square is 9.89949
时间复杂度: O(1)
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