Project Euler #6 - C

Question:

The sum of the squares of the first ten natural numbers is,

1² + 2² + ... + 10² = 385

The square of the sum of the first ten natural numbers is,

(1 + 2 + ... + 10)² = 55² = 3025

Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025  385 = 2640.

Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

Solution:

Maybe I am overlooking a snazzy shortcut or something, but this seems to be the most straightforward problem I have encountered on PE. Sum the squares, and square the sum . Take the difference, and be done with it. 

int main (void)
{
unsigned int sqr_o_sum = 0;
unsigned int sum_o_sqr = 0;
unsigned int difference;
int i;


for (i = 1; i <= 100; i++)
{
sum_o_sqr += i*i;
sqr_o_sum += i;
}
sqr_o_sum *= sqr_o_sum;
difference = sqr_o_sum - sum_o_sqr;


printf("Difference is %d\n", difference);


return 1;
}


Compile and watch the solution appear before your eyes!