Sum of primes below two million. Sieve of Eratosthenes

问题

Having a little trouble solving Problem: 'Calculate the sum of primes below two million'. I'm using the 'Sieve of Eratosthenes' method. My method works fine for finding primes till hundred but when I try to find the sum of primes till 2,000,000 I get an incorrect answer.

#include <iostream>

using namespace std;
long long unsigned int number[2000008];
int x=2000000LLU;
int sum()
{
    int s=0LLU; //stores sum
    for(int y=2; y<=x; y++) //add all the numers in the array from 2 to 2 million
    {
        s+=number[y];
    }
    return s;
}

int main()
{
    int k=2;
    for(int i=2; i<=x; i++) //fills in numbers from 2 to 2 million in the array
    {
        number[i]=i;
    }
    for(int j=2; j<=x; j+=1) //starts eliminating multiples of prime numbers from the grid
    {
        if(number[j]!=0) //moves through the grid till it finds a number that hasnt been crossed out. ie. isnt zero                            
        {
            for(int y=j+j; y<=x; y+=j) //when it finds a number, it removes all subsequent multiples of it
            {
                number[y]=0;
            }
        }

    }  
    cout<<endl<<"done"; //shows that the loop has been completed
    cout<<sum(); //outputs the sum of the grid
    return 0;
}

回答1:

I'm not sure an int is enough to hold the answer... It could be larger than a 32-bit value. Try using long long throughout.

回答2:

by using Sieve of Eratosthenes effectively, i solved the problem, here is my code , it is highly optimized

public class SumOfPrime {

    static void findSum()
    {
        long i=3;
        long sum=0;
        int count=0;
        boolean[] array = new boolean[2000000];
        for(long j=0;j<array.length;j++)
        {
         if((j&1)==0)
          array[(int)j]=false;   
         else    
         array[(int)j]=true;
        }
        array[1]=false;
        array[2]=true;
        for(;i<2000000;i+=2)
        { 
            if(array[(int)i] & isPrime(i))
            {   
                array[(int)i]=true;
                //Sieve of Eratosthenes
                for(long j=i+i;j<array.length;j+=i)
                    array[(int)j]=false;
            }
        }
        for(int j=0;j<array.length;j++)
        {
            if(array[j])
            {   
             //System.out.println(j);
             count++;   
             sum+=j;
            }
        }   
        System.out.println("Sum="+sum +" Count="+count);
    }
    public static boolean isPrime(long num)
    {
        boolean flag=false;
        long i=3;
        long limit=(long)Math.sqrt(num);
        for(;i<limit && !(flag);i+=2)
        {
            if(num%i==0)
            {
                flag=false;
                break;
            }   
        }
        if(i>=limit)
         flag=true;
        return flag;
    }

    public static void main(String args[])
    {
        long start=System.currentTimeMillis();
        findSum();
        long end=System.currentTimeMillis();
        System.out.println("Time for execution="+(end-start)+"ms");
    }

}

and the output is

Sum=142913828922 Count=148933
Time for execution=2360ms

if you have doubt, please do tell

来源：https://stackoverflow.com/questions/6176126/sum-of-primes-below-two-million-sieve-of-eratosthenes

标签

c++

sieve-of-eratosthenes