Tuesday 18 June 2013

PROJECT EULER SOLUTION # 1

Solution to problem number 1 of Project Euler.
Question # 1
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000.

/*****************************************************************************/
#include<stdio.h>
#include<conio.h>
#include<time.h>
int main()
{
                long sum=0;
                int i;
                for(i=0;i<1000;i++)
                                if(i%3==0 || i%5==0)
                                {
                                                sum+=i;
                                                printf("%d\t",i);
                                }
                printf("\n\n%ld\n",sum);
                printf("EXECUTION TIME = %f",clock()/(float)CLK_TCK);
                system("pause");

}
/*******************************************************************************/

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