Program to find sum of prime numbers between 1 to n
Write a program to find sum of all prime numbers between 1 to n.
Examples:
Input : 10
Output : 17
Explanation : Primes between 1 to 10 : 2, 3, 5, 7.
Input : 11
Output : 28
Explanation : Primes between 1 to 11 : 2, 3, 5, 7, 11.
A simple solution is to traverse all numbers from 1 to n. For every number, check if it is a prime. If yes, add it to result.
An efficient solution is to use Sieve of Eratosthenes to find all prime numbers from till n and then do their sum.
C++
#include <bits/stdc++.h>
using namespace std;
int sumOfPrimes( int n)
{
bool prime[n + 1];
memset (prime, true , n + 1);
for ( int p = 2; p * p <= n; p++) {
if (prime[p] == true ) {
for ( int i = p * 2; i <= n; i += p)
prime[i] = false ;
}
}
int sum = 0;
for ( int i = 2; i <= n; i++)
if (prime[i])
sum += i;
return sum;
}
int main()
{
int n = 11;
cout << sumOfPrimes(n);
return 0;
}
|
Java
import java.io.*;
import java.util.*;
class GFG {
static int sumOfPrimes( int n)
{
boolean prime[]= new boolean [n + 1 ];
Arrays.fill(prime, true );
for ( int p = 2 ; p * p <= n; p++) {
if (prime[p] == true ) {
for ( int i = p * 2 ; i <= n; i += p)
prime[i] = false ;
}
}
int sum = 0 ;
for ( int i = 2 ; i <= n; i++)
if (prime[i])
sum += i;
return sum;
}
public static void main(String args[])
{
int n = 11 ;
System.out.print(sumOfPrimes(n));
}
}
|
Python3
def sumOfPrimes(n):
prime = [ True ] * (n + 1 )
p = 2
while p * p < = n:
if prime[p] = = True :
i = p * 2
while i < = n:
prime[i] = False
i + = p
p + = 1
sum = 0
for i in range ( 2 , n + 1 ):
if (prime[i]):
sum + = i
return sum
n = 11
print (sumOfPrimes(n))
|
C#
using System;
class GFG {
static int sumOfPrimes( int n)
{
bool []prime= new bool [n + 1];
for ( int i = 0; i < n + 1; i++)
prime[i] = true ;
for ( int p = 2; p * p <= n; p++)
{
if (prime[p] == true )
{
for ( int i = p * 2; i <= n; i += p)
prime[i] = false ;
}
}
int sum = 0;
for ( int i = 2; i <= n; i++)
if (prime[i])
sum += i;
return sum;
}
public static void Main()
{
int n = 11;
Console.Write(sumOfPrimes(n));
}
}
|
PHP
<?php
function sumOfPrimes( $n )
{
$prime = array_fill (0, $n + 1, true);
for ( $p = 2;
$p * $p <= $n ; $p ++)
{
if ( $prime [ $p ] == true)
{
for ( $i = $p * 2;
$i <= $n ; $i += $p )
$prime [ $i ] = false;
}
}
$sum = 0;
for ( $i = 2; $i <= $n ; $i ++)
if ( $prime [ $i ])
$sum += $i ;
return $sum ;
}
$n = 11;
echo sumOfPrimes( $n );
?>
|
Javascript
<script>
function sumOfPrimes(n)
{
let prime = new Array(n + 1);
for (let i = 0; i < n + 1; i++)
prime[i] = true ;
for (let p = 2; p * p <= n; p++)
{
if (prime[p] == true )
{
for (let i = p * 2; i <= n; i += p)
prime[i] = false ;
}
}
let sum = 0;
for (let i = 2; i <= n; i++)
if (prime[i])
sum += i;
return sum;
}
let n = 11;
document.write(sumOfPrimes(n));
</script>
|
Output:
28
Time Complexity: O(nloglogn)
Auxiliary Space: O(n)
Last Updated :
30 Aug, 2022
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