Dado un valor n, encuentre la suma de la serie, (n/1) + (n/2) + (n/3) + (n/4) +…….+(n/n) donde el valor de n puede ser hasta 10^12.
Nota: Considere solo la división de enteros.
Ejemplos:
Input : n = 5 Output : (5/1) + (5/2) + (5/3) + (5/4) + (5/5) = 5 + 2 + 1 + 1 + 1 = 10 Input : 7 Output : (7/1) + (7/2) + (7/3) + (7/4) + (7/5) + (7/6) + (7/7) = 7 + 3 + 2 + 1 + 1 + 1 + 1 = 16
A continuación se muestra el programa para encontrar la suma de series dadas:
C++
// CPP program to find // sum of given series #include <bits/stdc++.h> using namespace std; // function to find sum of series long long int sum(long long int n) { long long int root = sqrt(n); long long int ans = 0; for (int i = 1; i <= root; i++) ans += n / i; ans = 2 * ans - (root * root); return ans; } // driver code int main() { long long int n = 35; cout << sum(n); return 0; }
Java
// Java program to find // sum of given series import java.util.*; class GFG { // function to find sum of series static long sum(long n) { long root = (long)Math.sqrt(n); long ans = 0; for (int i = 1; i <= root; i++) ans += n / i; ans = 2 * ans - (root * root); return ans; } /* Driver code */ public static void main(String[] args) { long n = 35; System.out.println(sum(n)); } } // This code is contributed by Arnav Kr. Mandal.
Python3
# Python 3 program to find # sum of given series import math # function to find sum of series def sum(n) : root = (int)(math.sqrt(n)) ans = 0 for i in range(1, root + 1) : ans = ans + n // i ans = 2 * ans - (root * root) return ans # driver code n = 35 print(sum(n)) # This code is contributed by Nikita Tiwari.
C#
// C# program to find // sum of given series using System; class GFG { // Function to find sum of series static long sum(long n) { long root = (long)Math.Sqrt(n); long ans = 0; for (int i = 1; i <= root; i++) ans += n / i; ans = 2 * ans - (root * root); return ans; } // Driver code public static void Main() { long n = 35; Console.Write(sum(n)); } } // This code is contributed vt_m.
PHP
<?php // PHP program to find // sum of given series // function to find // sum of series function sum($n) { $root = intval(sqrt($n)); $ans = 0; for ($i = 1; $i <= $root; $i++) $ans += intval($n / $i); $ans = (2 * $ans) - ($root * $root); return $ans; } // Driver code $n = 35; echo (sum($n)); // This code is contributed by // Manish Shaw(manishshaw1) ?>
Javascript
<script> // Javascript program to find // sum of given series // function to find // sum of series function sum(n) { let root = parseInt(Math.sqrt(n)); let ans = 0; for (let i = 1; i <= root; i++) ans += parseInt(n / i); ans = (2 * ans) - (root * root); return ans; } // Driver code let n = 35; document.write(sum(n)); // This code is contributed by gfgking. </script>
Producción:
131
Nota: Si se observa de cerca, podemos ver que, si tomamos n común, la serie se convierte en una Progresión Armónica .