Project Euler #1: Multiples 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.

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

Constraints
1≤T≤105
1≤N≤109

Sample Input

2
10
100

Sample Output

23
2318 

Language: C

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#include <stdio.h>

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#include <stdlib.h>

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​

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int main()

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{

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    unsigned long long int N,i,j,x,y,z;

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    int T;

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    scanf("%d",&T);

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    unsigned long long int n[T];

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    for(i=1;i<=T;i++)

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    {

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        scanf("%llu",&N);

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        x=(N-1)/3;

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        y=(N-1)/5;

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        z=y/3;

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        n[i-1]=(3+3*x)*x/2;

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        n[i-1]+=(5+5*y)*y/2;

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        n[i-1]-=(15+15*z)*z/2;

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    }

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    for(i=1;i<=T;i++)

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    {

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        printf("%llu\n",n[i-1]);

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    }

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    return 0;

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}