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 1000.


Can someone give me suggestions on how to improve this solution? Also, please tell me whether or not my programming style is good.

#include&lt;stdio.h&gt;

int main()
{
    int a,b,c,d,sum3,sum5,sum15,total_sum;
    b=999/3;
    c=999/5;
    d=999/15;
    sum3=b*(3+999)/2;
    sum5=c*(5+995)/2;
    sum15=d*(15+990)/2;
    total_sum=sum3+sum5-sum15;
    printf("%d",total_sum);
    return 0;
}
Functions in C with no parameters should have a void parameter:

int main(void)

Variables should be declared/initialized on separate lines:

int a;
int b;
int c;
int d;
int sum3;
int sum5;
int sum15;
int total_sum;

Try to avoid single-character variable names.  They're very ambiguous and say nothing about the variable's purpose.  The only valid exception to this is loop counter variables.
Statements/blocks of different purposes should be separated to increase readability.  That would be the divisions, the sums, and the printf().
Your operands should be spread out to make the lines more readable:

sum3 = b * (3+999) / 2;

Initializations should be done as late as possible to help aid in maintenance:

sum3 = b * (3+999) / 2;

Remove int a since it's never used.