Project Euler #1: Multiples of 3 and 5 | Hackerrank | Project Euler

Multiple of 3 and 5 

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 N.

Input Format

First line contains T that denotes the number of test cases. This is followed by T lines, each containing an integer, N.

Constraints

1<=T<=10^5

1<=T<=10^9

Output Format

For each test case, print an integer that denotes the sum of all the multiples of  3 or 5 below  N.

Sample Input 0

2

10

100

Sample Output 0

23

2318

Explanation 0

For N=10, 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.

Similarly for N=100, we get 2318.

Solution:

#include <math.h>

#include <stdio.h>

#include <string.h>

#include <stdlib.h>

#include <assert.h>

#include <limits.h>

#include <stdbool.h>

int main(){

    int t;

    long int s3,s5,s15,f3,f5,f15; 

    scanf("%d",&t);

    for(int a0 = 0; a0 < t; a0++){

        long int sum=0,n;

        scanf("%ld",&n);

        f3=(n-1)/3;

        f5=(n-1)/5;

        f15=(n-1)/15;

        s3=3*((f3*(f3+1))/2);

        s5=5*((f5*(f5+1))/2);

        s15=15*((f15*(f15+1))/2);

        sum=s5+s3-s15;

        printf("%ld\n",sum);

    }

    return 0;

}

You can also run it on an online IDE:   

https://ide.geeksforgeeks.org/f83046d7-4cfa-4f66-a911-107e1c89e14b

Your feedback are always welcome! If you have any doubt you can contact me or leave a comment!  Happy Coding !! Cheers!!!

Related Links:

https://codepiggy.blogspot.com/2021/02/count-pairs-code-vita-9-zonal-round.html