求 A 和 B 之间的 N 个几何平均数
原文:https://www . geesforgeks . org/find-n-geometric-means-介于-a 和-b 之间/
给定三个整数 A、B 和 N，任务是找到 A 和 B 之间的 N 个几何平均值。我们基本上需要在一个几何级数中插入 N 个项。其中 A 和 B 是第一项和最后一项。 例:
Input : A = 2 B = 32 N = 3
Output : 4 8 16
the geometric progression series as 2,
4, 8, 16 , 32
Input : A = 3 B = 81 N = 2
Output : 9 27
进场: 让 A 1 ，G 2 ，G 3 ，G 4 ……G n 为两个给定数字 A 和 B 之间的 N 个几何平均数，然后 A、G 1 ，G 2 …..G n ，B 将处于几何级数。 所以 B = (N+2) 几何级数的第项。 那么这里 R 是公比 B = A * RN+1RN+1= B/A T30】R =(B/A)1/(N+1)T34】现在我们有了 R 的值，也有了第一项 A G 1 的值= AR (B/A)2/(N+1)T50】G3= AR3= A (B/A)3/(N+1) 。 。 。 GN= ARN= A (B/A)N/(N+1)**
C++
// C++ program to find n geometric means
// between A and B
#include <bits/stdc++.h>
using namespace std;
// Prints N geometric means between
// A and B.
void printGMeans(int A, int B, int N)
{
// calculate common ratio(R)
float R = (float)pow(float(B / A),
1.0 / (float)(N + 1));
// for finding N the Geometric
// mean between A and B
for (int i = 1; i <= N; i++)
cout << A * pow(R, i) <<" ";
}
// Driver code to test above
int main()
{
int A = 3, B = 81, N = 2;
printGMeans(A, B, N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// java program to illustrate
// n geometric mean between
// A and B
import java.io.*;
import java.lang.*;
import java.util.*;
public class GFG {
// insert function for calculating the means
static void printGMeans(int A, int B, int N)
{
// Finding the value of R Common ration
float R = (float)Math.pow((float)(B / A),
1.0 / (float)(N + 1));
// for finding N the Geometric
// mean between A and B
for (int i = 1; i <= N; i++)
System.out.print(A * Math.pow(R, i) + " ");
}
// Driver code
public static void main(String args[])
{
int A = 3, B = 81, N = 2;
printGMeans(A, B, N);
}
}
Python 3
# Python3 program to find
# n geometric means
# between A and B
import math
# Prints N geometric means
# between A and B.
def printGMeans(A, B, N):
# calculate
# common ratio(R)
R = (math.pow((B / A),
1.0 / (N + 1)));
# for finding N the
# Geometric mean
# between A and B
for i in range(1, N + 1):
print(int(A * math.pow(R, i)),
end = " ");
# Driver Code
A = 3;
B = 81;
N = 2;
printGMeans(A, B, N);
# This code is contributed
# by mits
C
// C# program to illustrate
// n geometric mean between
// A and B
using System;
public class GFG {
// insert function for calculating the means
static void printGMeans(int A, int B, int N)
{
// Finding the value of R Common ration
float R = (float)Math.Pow((float)(B / A),
1.0 / (float)(N + 1));
// for finding N the Geometric
// mean between A and B
for (int i = 1; i <= N; i++)
Console.Write(A * Math.Pow(R, i) + " ");
}
// Driver code
public static void Main()
{
int A = 3, B = 81, N = 2;
printGMeans(A, B, N);
}
}
// 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 geometric means
// between A and B
// Pr$s N geometric means
// between A and B.
function printGMeans($A, $B, $N)
{
// calculate common ratio(R)
$R = pow(($B / $A),
1.0 / ($N + 1));
// for finding N the Geometric
// mean between A and B
for ($i = 1; $i <= $N; $i++)
echo $A * pow($R, $i) ," ";
}
// Driver Code
$A = 3;
$B = 81;
$N = 2;
printGMeans($A, $B, $N);
// This code is contributed by anuj_67.
?>
java 描述语言
<script>
// JavaScript program to illustrate
// n geometric mean between
// A and B
// insert function for calculating the means
function printGMeans(A, B, N)
{
// Finding the value of R Common ration
let R = Math.pow((B / A),
1.0 / (N + 1));
// for finding N the Geometric
// mean between A and B
for (let i = 1; i <= N; i++)
document.write(A * Math.pow(R, i) + " ");
}
// Driver Code
let A = 3, B = 81, N = 2;
printGMeans(A, B, N);
// This code is contributed by code_hunt.
</script>
输出:
9 27
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