求A和B之间的N个几何平均数

给定三个整数A、B和N,任务是在A和B之间找到N个几何平均值。我们基本上需要在A中插入N个项 几何级数 .其中A和B是第一项和最后一项。 例如:

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Input :  A = 2  B = 32  N = 3Output : 4 8 16the geometric progression series as 2,4, 8, 16 , 32       Input : A = 3 B = 81 N = 2Output : 9 27

方法: 让我们 1. G 2. G 3. G 4. ……G N N是两个给定数字A和B之间的几何平均数。然后A,G 1. G 2. ….. G N ,B将是几何级数。 所以B=(N+2) th 几何级数的术语。 这里R是公共比率 B=A*R N+1 R N+1 =B/A R=(B/A) 1/(N+1) 现在我们有了R的值 还有第一项A的值 G 1. =AR 1. =A*(B/A) 1/(N+1) G 2. =AR 2. =A*(B/A) 2/(N+1) G 3. =AR 3. =A*(B/A) 3/(N+1) . . . G N =AR N =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

// 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);
}
}


Python3

# 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.


PHP

<?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.
?>


Javascript

<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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