Project Euler #1 test extension - c++

Question

The simple solution to problem 1 is
static unsigned int solutionInefficient(unsigned int n){
unsigned int sum = 0;
for (unsigned int i = 0; i < n; i++){
if (i % 3 == 0 || i % 5 == 0) {
sum += i;
}
}
return sum;
}
I decided to try a different test case with n = 2147483647 and the final result was computed in 12 seconds. So, I came up with another solution that gave me the same answer and took 2 seconds:
static unsigned int solutionEfficient(unsigned int n){
unsigned int sum = 0;
unsigned int sum3 = 0;
unsigned int sum5 = 0;
unsigned int sum15 = 0;
for (unsigned int i = 3; i < n; i += 3){
sum3 += i;
}
for (unsigned int i = 5; i < n; i += 5){
sum5 += i;
}
for (unsigned int i = 15; i < n; i += 15){
sum15 += i;
}
return sum3 + sum5 - sum15;
}
My last attempt at making a faster implementation involved some google searches and using the arithmetic summation formula and the final piece of code looked like this:
static unsigned int solutionSuperEfficient(unsigned int n){
n = n - 1;
unsigned int t3 = n / (unsigned int)3,
t5 = n / (unsigned int)5,
t15 = n / (unsigned int)15;
unsigned int res_3 = 3 * (t3 * (t3 + 1)) *0.5;
unsigned int res_5 = 5 * (t5 * (t5 + 1)) *0.5;
unsigned int res_15 = 15 * (t15 * (t15 + 1)) *0.5;
return res_3 + res_5 - res_15;
}
however this did not provide the correct answer for this test case. It did provide the correct answer for n = 1000. I am not sure why it failed for my test case, any ideas?

Answer 1

You have two problems in your super efficient solution:
You are using floating point number 0.5 instead of dividing by 2. This will cause rounding errors. Note that it is guaranteed that x * (x + 1) is even so you can safely divide by two.
Integer overflow. The calculation t3 * (t3 + 1) and the similar products will overflow unsigned int. To avoid this, use unsigned long long instead.
Here is the corrected code:
static unsigned int solutionSuperEfficient(unsigned int n){
n = n - 1;
unsigned long long t3 = n / 3,
t5 = n / 5,
t15 = n / 15;
unsigned long long res_3 = 3ULL * ((t3 * (t3 + 1)) / 2ULL);
unsigned long long res_5 = 5ULL * ((t5 * (t5 + 1)) / 2ULL);
unsigned long long res_15 = 15LL * ((t15 * (t15 + 1)) / 2ULL);
return (unsigned int)(res_3 + res_5 - res_15);
}
In fact you don't need t3, t5 and t15 to be unsigned long long as those values would never overflow unsigned int.