你得到了一系列的1/1!+2/2! + 3/3! + 4/4! +…….+ 不!,求出直到第n项的级数之和。 例如:
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Input :n = 5Output : 2.70833Input :n = 7Output : 2.71806
A. 简单解决方案 是一个接一个的计算机术语。对于每个项,找到其对应的阶乘值。 一 有效解决方案 就是在同一个循环中进行阶乘计算。
CPP
// CPP program to print // the sum of series #include<bits/stdc++.h> using namespace std; // function to calculate // sum of given series double sumOfSeries( double num) { double res = 0, fact = 1; for ( int i = 1; i <= num; i++) { // fact variable store // factorial of the i fact = fact * i; res = res + (i / fact); } return (res); } // Driver Code int main() { double n = 5; cout << "Sum: " << sumOfSeries(n); return 0; } |
JAVA
// Java program to print // the sum of series import java.io.*; import java.lang.*; class GFG { public static double sumOfSeries( double num) { double res = 0 , fact = 1 ; for ( int i = 1 ; i <= num; i++) { // fact variable store // factorial of the i fact = fact * i; res = res + (i / fact); } return (res); } // Driver Code public static void main (String[] args) { double n = 5 ; System.out.println( "Sum: " + sumOfSeries(n)); } } // This code is contributed by // Mohit Gupta_OMG <(0_o)> |
Python3
# Python code to find smallest K-digit # number divisible by X def sumOfSeries(num): # Computing MAX res = 0 fact = 1 for i in range ( 1 , num + 1 ): fact * = i res = res + (i / fact) return res n = 5 print ( "Sum: " , sumOfSeries(n)) # Code contributed by # Mohit Gupta_OMG <(0_o)> |
C#
// C# program to print the sum of series using System; class GFG { public static float sumOfSeries( double num) { float res = 0, fact = 1; for ( int i = 1; i <= num; i++) { // fact variable store // factorial of the i fact = fact * i; res = res + (i / fact); } return (res); } // Driver Code public static void Main () { double n = 5; Console.Write( "Sum: " + sumOfSeries(n)); } } // This code is contributed by vt_m. |
PHP
<?php // PHP program to print // the sum of series // Function to calculate // sum of given series function sumOfSeries( $num ) { $res = 0; $fact = 1; for ( $i = 1; $i <= $num ; $i ++) { // fact variable store // factorial of the i $fact = $fact * $i ; $res = $res + ( $i / $fact ); } return ( $res ); } // Driver Code $n = 5; echo ( "Sum: " . sumOfSeries( $n )); // This code is contributed by Ajit. ?> |
Javascript
<script> // javascript program to print // the sum of series function sumOfSeries(num) { var res = 0, fact = 1; for (i = 1; i <= num; i++) { // fact variable store // factorial of the i fact = fact * i; res = res + (i / fact); } return (res); } // Driver Code var n = 5; document.write( "Sum: " + sumOfSeries(n).toFixed(5)); // This code is contributed by Amit Katiyar </script> |
输出:
Sum: 2.70833
时间复杂度:O(n)
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