给定N和标准偏差，找到N个元素

给定N和标准差，则找到N个元素。

Mean is average of element. 
Mean of arr[0..n-1] = Σ(arr[i]) / n 
where 0 <= i < n
Variance is sum of squared differences from the mean divided by number of elements.
Variance = Σ(arr[i] – mean)² / n
Standard Deviation is square root of variance 
Standard Deviation = Σ(variance)
Please refer Mean, Variance and Standard Deviation for details. 
 

例子：

Input: 6 0 
Output: 0 0 0 0 0 0  
Explanation: 
The standard deviation of 0, 0, 0, 0, 0, 0 is 0. 
Also the standard deviation of 4, 4, 4, 4, 4, 4 
is 0, we print any of the possible N elements.

Input: 3 3
Output: 0 -3.67423 3.67423 
Explanation: 
On calculating SD of these N elements,
we get standard deviation to be 3. 
            

方法：
如果看一下公式，我们有两个未知项，一个是xi，另一个是均值。主要动机是使平均值为0，以便我们可以获取X元素的公式。将有两种情况，一种是偶数，一种是奇数。

当N为偶数时：
要使N个元素的平均值为0，最好的方法是将N个元素表示为-X + X -X + X…。公式将是sqrt((x ^ 2)/ n的总和)，x2 + x ^ 2 + x ^ 2 +………N个项，因此公式将是sqrt(N *(x ^ 2)/ N) 。 N互相抵消，因此sqrt(x ^ 2)变成SD。因此，我们得到N个元素为-SD + SD -SD + SD……以得到均值0。我们需要打印-SD + SD -SD + SD……

当N为奇数时：
N个元素的平均值为0。因此，一个元素为0，其他N-1个元素为-X + X -X…。公式将为sqrt((x ^ 2)/ n的总和)，x2 + x ^ 2 + x ^ 2 + ………N-1个项，因此公式为sqrt((N-1)*(x ^ 2)/ N)，所以
X = SD * sqrt(n /(n-1))。 n个元素是0 -X + X -X + X…
当SD为0时，所有元素都将相同，因此我们可以为其打印0。

下面是上述方法的实现：

C++

// CPP program to find n elements
#include 
using namespace std;
 
// function to print series of n elements
void series(int n, int d)
{
 
    // if S.D. is 0 then print all
    // elements as 0.
    if (d == 0) {
 
        // print n 0's
        for (int i = 0; i < n; i++)
            cout << "0 ";
 
        cout << endl;
        return;
    }
 
    // if S.D. is even
    if (n % 2 == 0) {
 
        // print -SD, +SD, -SD, +SD
        for (int i = 1; i <= n; i++) {
            cout << pow(-1, i) * d << " ";
        }
        cout << endl;
    }
    else // if odd
    {
        // convert n to a float integer
        float m = n;
        float r = (m / (m - 1));
        float g = (float)(d * (float)sqrtf(r));
 
        // print one element to be 0
        cout << "0 ";
 
        // print (n-1) elements as xi derived
        // from the formula
        for (int i = 1; i < n; i++) {
            cout << pow(-1, i) * g << " ";
        }
        cout << endl;
    }
}
 
// driver program to test the above function
int main()
{
    int n = 3, d = 3;
    series(n, d);
 
    return 0;
}

Java

// Java program to find n elements
import java.util.*;
import java.lang.*;
 
public class GfG {
 
    // function to print series of n elements
    public static void series(int n, int d)
    {
 
        // if S.D. is 0 then print all
        // elements as 0.
        if (d == 0) {
 
            // print n 0's
            for (int i = 0; i < n; i++)
                System.out.print("0 ");
            System.out.println();
            return;
        }
 
        // if S.D. is even
        if (n % 2 == 0) {
 
            // print -SD, +SD, -SD, +SD
            for (int i = 1; i <= n; i++) {
                System.out.print(Math.pow(-1, i) * d + " ");
            }
            System.out.println();
        }
        else // if odd
        {
            // convert n to a float integer
            float m = n;
            float r = (m / (m - 1));
            float g = (float)(d * (float)(Math.sqrt(r)));
 
            // print one element to be 0
            System.out.print("0 ");
 
            // print (n-1) elements as xi
            // derived from the formula
            for (int i = 1; i < n; i++) {
                System.out.print(Math.pow(-1, i) * g + " ");
            }
            System.out.println();
        }
    }
 
    // driver function
    public static void main(String args[])
    {
        int n = 3, d = 3;
        series(n, d);
    }
}
 
/* This code is contributed by Sagar Shukla */

Python3

# Python program to find n elements
import math
 
# function to print series of n elements
def series( n, d):
 
    # if S.D. is 0 then print all
    # elements as 0.
    if d == 0:
     
        # print n 0's
        for i in range(n):
            print("0", end = ' ')
        return 1
         
    # if S.D. is even
    if n % 2 == 0:
     
        # print -SD, +SD, -SD, +SD
        i = 1
        while i <= n:
            print("%.5f"%((math.pow(-1, i) * d)),
                  end =' ')
            i += 1
    else:
        # if odd
        # convert n to a float integer
        m = n
        r = (m / (m - 1))
        g = (float)(d * float(math.sqrt(r)))
         
        # print one element to be 0
        print("0 ", end = ' ')
         
        # print (n-1) elements as xi derived
        # from the formula
        i = 1
        while i < n:
            print("%.5f"%(math.pow(-1, i) * g),
                  end = ' ')
            i = i + 1
    print("\n")
 
# driver code to test the above function
n = 3
d = 3
series(n, d)
 
# This code is contributed by "Sharad_Bhardwaj".

C#

// C# program to find n elements
using System;
 
public class GfG {
 
    // function to print series of n
    // elements
    public static void series(int n, int d)
    {
 
        // if S.D. is 0 then print all
        // elements as 0.
        if (d == 0) {
 
            // print n 0's
            for (int i = 0; i < n; i++)
                Console.Write("0");
                 
            Console.WriteLine();
             
            return;
        }
 
        // if S.D. is even
        if (n % 2 == 0) {
 
            // print -SD, +SD, -SD, +SD
            for (int i = 1; i <= n; i++) {
                Console.Write(Math.Pow(-1, i)
                                   * d + " ");
            }
             
            Console.WriteLine();
        }
        else // if odd
        {
             
            // convert n to a float integer
            float m = n;
            float r = (m / (m - 1));
            float g = (float)(d *
                       (float)(Math.Sqrt(r)));
 
            // print one element to be 0
            Console.Write("0 ");
 
            // print (n-1) elements as xi
            // derived from the formula
            for (int i = 1; i < n; i++) {
                Console.Write(Math.Pow(-1, i)
                                   * g + " ");
            }
             
            Console.WriteLine();
        }
    }
 
    // driver function
    public static void Main()
    {
        int n = 3, d = 3;
         
        series(n, d);
    }
}
 
// This code is contributed by vt_m.

Javascript

 
// JavaScript program to find n elements
 
// Function to prlet series of n elements
function series(n, d)
{
     
    // If S.D. is 0 then prlet all
    // elements as 0.
    if (d == 0)
    {
         
        // Print n 0's
        for(let i = 0; i < n; i++)
            document.write("0 ");
             
        document.write();
        return;
    }
 
    // If S.D. is even
    if (n % 2 == 0)
    {
         
        // Print -SD, +SD, -SD, +SD
        for(let i = 1; i <= n; i++)
        {
            document.write(Math.pow(-1, i) * d + "  ");
        }
        document.write();
    }
     
    // If odd
    else
    {
         
        // Convert n to a float leteger
        let m = n;
        let r = (m / (m - 1));
        let g = (d * (Math.sqrt(r)));
 
        // Print one element to be 0
        document.write("0 ");
 
        // Print (n-1) elements as xi
        // derived from the formula
        for(let i = 1; i < n; i++)
        {
            document.write(Math.pow(-1, i) * g + "  ");
        }
        document.write();
    }
}
 
// Driver Code
let n = 3, d = 3;
 
series(n, d);
 
// This code is contributed by susmitakundugoaldanga
 

输出：

0 -3.67423 3.67423