Consider the following C++ procedure.

int sum(long long n)
{
    int result = 0;
    while (n) {
        result += n % 10;
        n /= 10;
    }
    return result;
}

long long foo(long long n, int target)
{
    long long x = n, y = 1;
    while (sum(n) > target) {
        n = n / 10 + 1;
        y *= 10;
    }
    return n * y - x;
}