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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