给定 N 和标准差，找到 N 个元素
原文:https://www . geesforgeks . org/given-n-标准差-find-n-elements/
给定 N 和标准差求 N 个元素。
平均值是元素的平均值。 表示 arr【0】的..n-1]=σ(arr[I])/n 其中 0 < = i < n 方差是平均值除以元素数的平方差之和。 方差=σ(arr[I]–均值) 2 / n 标准差为方差的平方根 标准差=σ(方差) 详见均值、方差和标准差。
示例:
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\.
进场: 如果我们看公式，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，这样我们就可以得到 X 元素的公式。会有两种情况，一种是偶数，一种是奇数。
当 N 为偶数时: 要使 N 元素的平均值为 0，最好的方法是将 N 元素表示为-X +X -X +X。公式将是 sqrt(x^2)/n、x2+x^2+x^2+………N 项之和)，所以公式原来是 sqrt (N(x^2)/N).n 相互抵消，所以 sqrt (x^2)变成了 SD。所以，我们得到 N 元素为 -SD +SD -SD +SD……* 得到平均值 0。我们需要打印-标清+标清-标清+标清……
N 为奇数时: N 元素的平均值为 0。所以，一个元素是 0，其他 N-1 个元素是-X +X -X。公式将是 sqrt(x^2)/n、x2+x^2+x^2+………N-1 项之和，所以公式原来是 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，那么所有元素都是相同的，所以我们可以为其打印 0。
下面是上述方法的实现:
C++
// CPP program to find n elements
#include <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)
// 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 */
Python 3
# 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.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find n elements
// function to print
// series of n elements
function series( $n, $d)
{
// if S.D. is 0 then print all
// elements as 0.
if ($d == 0)
{
// print n 0's
for ($i = 0; $i < $n;$i++)
echo "0 ";
echo "\n";
return;
}
// if S.D. is even
if ($n % 2 == 0)
{
// print -SD, +SD, -SD, +SD
for ( $i = 1; $i <= $n; $i++)
{
echo pow(-1, $i) * $d , " ";
}
echo"\n";
}
// if odd
else
{
// convert n to a
// float integer
$m = $n;
$r = ($m / ($m - 1));
$g = ($d * sqrt($r));
// print one element
// to be 0
echo "0 ";
// print (n-1) elements
// as xi derived
// from the formula
for ($i = 1; $i < $n; $i++)
{
echo pow(-1, $i) * $g , " ";
}
echo"\n";
}
}
// Driver Code
$n = 3; $d = 3;
series($n, $d);
// This code is contributed by anuj_67.
?>
java 描述语言
<script>
// JavaScript program to find n elements
// Function to print series of n elements
function series(n, d)
{
// If S.D. is 0 then print 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 integer
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
</script>
输出:
0 -3.67423 3.67423
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