PROJECT EULER SOLUTION # 50

Solution to problem number 50 of Project Euler.

Question # 50

The prime 41, can be written as the sum of six consecutive primes:

41 = 2 + 3 + 5 + 7 + 11 + 13

This is the longest sum of consecutive primes that adds to a prime below one-hundred.

The longest sum of consecutive primes below one-thousand that adds to a prime, contains 21 terms, and is equal to 953.

Which prime, below one-million, can be written as the sum of the most consecutive primes?

Solution # 50

/***********************************************************************************************************/

#include<stdio.h>

#include<conio.h>

#include<stdlib.h>

#include<time.h>

#include<math.h>

int is_prime(long);

int is_prime(long n)

{

       int i=2;

       for(i=2;i<=(long)sqrt((double)n);i++)

              if(n%i==0)

                     return 0;

       return 1;

}

int main()

{

       long i,j,sum=0,temp_terms=0,terms=0,ans=0,temp,num=1000,a,actual_sum=0;

       for(i=2;i<num && ans<num;i++)

       {

              if(is_prime(i))

              {

                     sum=0;

                     temp_terms=0;

                     a=0;

                     for(j=i;;j++)

                     {

                          

                           temp=sum;

                    

                           if(is_prime(j))

                           {

                                  a++; // a counts the sequence of primes...

                                  sum+=j;

                                  if(sum>=num)

                                         break;

                                  if(is_prime(sum))

                                  {

                                         actual_sum=sum;

                                         temp_terms=a;

                                  }

                           }

                     }

                     if(temp_terms>terms)

                     {

                           terms=temp_terms;

                           ans=actual_sum;

                     }

              }

       }

       printf("%d\n",ans);

       printf("Execution time = %f\n",clock()/(float)CLK_TCK);

       system("pause");

}

/***********************************************************************************************************/