Missing & Duplicate

int Source[]; // The initial array.

// First, we calculate the difference between the sum of all the elements of the array and the sum of the integers from 1 to N.

int sum_ref = 0;
int sum_source = 0;

int index = 0;

foreach (int i in Source)
{
    sum_ref += ++index;
    sum_source += i;
}

int diff = abs(sum_source - sum_ref);

// Now we build two sums : Sum1 (resp. Sum2) is the sum of all the integers which euclidean division by diff is even (resp. odd).
// (Once again, ref is for integers 1 to N and source is for the Array)

int sum1_ref = 0;
int sum1_source = 0;
int sum2_ref = 0;
int sum2_source = 0;

index = 0;

foreach (int i in Source)
{
    index++;
    if (isEven(div(index, diff)))
        sum1_ref += index
    else
        sum2_ref += index;

    if (isEven(div(i, diff)))
        sum1_source += i
    else
        sum2_source += i;
}

// Finally, the positive difference between ref and source is the missing number, and the negative one is the duplicate.

int duplicate, missing;

if (sum1_ref > sum1_source)
{
    missing = sum1_ref - sum1_source;
    duplicate = sum2_source - sum2_ref;
}
else
{
    missing = sum2_ref - sum2_source;
    duplicate = sum1_source - sum1_ref;
}