HACKER RANK Project Euler 01 - Multiples of 3 and 5 - Solution.'C'

Project Euler Challenges-01 - Multiples of 3 and 5 Solution

Problem Statement

This problem is a programming version of Problem 1 from projecteuler.net

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.

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

Constraints
1≤T≤105
1≤N≤109

Sample Input

2
10
100

Sample Output

23
2318

Solution:

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {
unsigned long long n,i,sum=0,t;
    scanf("%llu",&t);
    while(t--)
        {
   
        scanf("%lld",&n);
        --n;
         unsigned long long n1,n2,n3;
         n1=n/3;
         n2=n/5;
         n3=n/15;
         sum=(3*n1*(n1+1))/2+(5*n2*(n2+1))/2-(15*n3*(n3+1))/2;
         printf("%lld\n",sum);
    }
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */    
    return 0;
}

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