前 n 个奇数自然数的平均值
原文:https://www . geesforgeks . org/average-of-first-n-奇数-naturals-numbers/
给定一个数 n，然后求前 n 个奇数的平均值 1 + 3 + 5 + 7 + 9 +…………。+(2n–1) 例:
Input : 5
Output : 5
(1 + 3 + 5 + 7 + 9)/5 = 5
Input : 10
Output : 10
(1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19)/10 =10
方法 1(天真方法: 一个简单的解决方案是迭代循环形式 1 到 n 次。通过所有奇数的和并除以 N，这个解需要 O(N)个时间。
C++
// A C++ program to find average of
// sum of first n odd natural numbers.
#include <iostream>
using namespace std;
// Returns the Avg of
// first n odd numbers
int avg_of_odd_num(int n)
{
// sum of first n odd number
int sum = 0;
for (int i = 0; i < n; i++)
sum += (2 * i + 1);
// Average of first
// n odd numbers
return sum / n;
}
// Driver Code
int main()
{
int n = 20;
cout << avg_of_odd_num(n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find average of
// sum of first n odd natural numbers.
import java.io.*;
class GFG {
// Returns the Avg of
// first n odd numbers
static int avg_of_odd_num(int n)
{
// sum of first n odd number
int sum = 0;
for (int i = 0; i < n; i++)
sum += (2 * i + 1);
// Average of first
// n odd numbers
return sum / n;
}
// Driver Code
public static void main(String[] args)
{
int n = 20;
avg_of_odd_num(n);
System.out.println(avg_of_odd_num(n));
}
}
// This code is contributed by vt_m
Python 3
# A Python 3 program
# to find average of
# sum of first n odd
# natural numbers.
# Returns the Avg of
# first n odd numbers
def avg_of_odd_num(n) :
# sum of first n odd number
sm = 0
for i in range(0, n) :
sm = sm + (2 * i + 1)
# Average of first
# n odd numbers
return sm//n
# Driver Code
n = 20
print(avg_of_odd_num(n))
# This code is contributed
# by Nikita Tiwari.
C
// C# program to find average
// of sum of first n odd
// natural numbers.
using System;
class GFG {
// Returns the Avg of
// first n odd numbers
static int avg_of_odd_num(int n)
{
// sum of first n odd number
int sum = 0;
for (int i = 0; i < n; i++)
sum += (2 * i + 1);
// Average of first
// n odd numbers
return sum / n;
}
// Driver code
public static void Main()
{
int n = 20;
avg_of_odd_num(n);
Console.Write(avg_of_odd_num(n));
}
}
// This code is contributed by
// Smitha Dinesh Semwal
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// A PHP program to find average of
// sum of first n odd natural numbers.
// Returns the Avg of
// first n odd numbers
function avg_of_odd_num($n)
{
// sum of first n odd number
$sum = 0;
for ($i = 0; $i < $n; $i++)
$sum += (2 * $i + 1);
// Average of first
// n odd numbers
return $sum / $n;
}
// Driver Code
$n = 20;
echo(avg_of_odd_num($n));
// This code is contributed by Ajit.
?>
java 描述语言
<script>
// javascript program to find average of
// sum of first n odd natural numbers.
// Returns the Avg of
// first n odd numbers
function avg_of_odd_num( n)
{
// sum of first n odd number
let sum = 0;
for (let i = 0; i < n; i++)
sum += (2 * i + 1);
// Average of first
// n odd numbers
return sum / n;
}
// Driver Code
let n = 20;
document.write(avg_of_odd_num(n));
// This code is contributed by todaysgaurav
</script>
输出:
20
时间复杂度:O(n) 方法二(有效途径: 思路是前 n 个奇数的和为 n 2 ，求前 n 个奇数的平均值，所以除以 n，因此公式为 n 2 /n = n 。这需要时间。
Avg of sum of first N odd Numbers = N
C++
// CPP Program to find the average
// of sum of first n odd numbers
#include <bits/stdc++.h>
using namespace std;
// Return the average of sum
// of first n odd numbers
int avg_of_odd_num(int n)
{
return n;
}
// Driver Code
int main()
{
int n = 8;
cout << avg_of_odd_num(n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// java Program to find the average
// of sum of first n odd numbers
import java.io.*;
class GFG {
// Return the average of sum
// of first n odd numbers
static int avg_of_odd_num(int n)
{
return n;
}
// Driver Code
public static void main(String[] args)
{
int n = 8;
System.out.println(avg_of_odd_num(n));
}
}
// This code is contributed by vt_m
Python 3
# Python 3 Program to
# find the average
# of sum of first n
# odd numbers
# Return the average of sum
# of first n odd numbers
def avg_of_odd_num(n) :
return n
# Driver Code
n = 8
print(avg_of_odd_num(n))
# This code is contributed
# by Nikita Tiwari.
C
// C# Program to find the average
// of sum of first n odd numbers
using System;
class GFG {
// Return the average of sum
// of first n odd numbers
static int avg_of_odd_num(int n)
{
return n;
}
// Driver Code
public static void Main()
{
int n = 8;
Console.Write(avg_of_odd_num(n));
}
}
// This code is contributed by
// Smitha Dinesh Semwal
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP Program to find the average
// of sum of first n odd numbers
// Return the average of sum
// of first n odd numbers
function avg_of_odd_num($n)
{
return $n;
}
// Driver Code
$n = 8;
echo(avg_of_odd_num($n));
// This code is contributed by Ajit.
?>
java 描述语言
<script>
// javascript Program to find the average
// of sum of first n odd numbers
// Return the average of sum
// of first n odd numbers
function avg_of_odd_num(n)
{
return n;
}
// Driver Code
var n = 8;
document.write(avg_of_odd_num(n));
// This code is contributed by gauravrajput1
</script>
输出:
8
时间复杂度:O(1) 证明
Sum of first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series
and d is the difference between the adjacent
terms of the series.
Here, a = 1, d = 2, applying these values to e. q.,
(i), we get
Sum = (n/2) * [2*1 + (n-1)*2]
= (n/2) * [2 + 2*n - 2]
= (n/2) * (2*n)
= n*n
= n2
Avg of first n odd numbers = n2/n
= n
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