How can I make this function output max array value? - c++

Question

I'm new to C++ and trying to make this piece of code work properly. It's a basic algorithm for
prime numbers. I make an array the size of which is defined by the user input. Then I need to print
only the max prime number from this array. How can I do that here?
unsigned long long nPrime(unsigned n) {
long long *a = new long long[n + 1];
long long maxNum = a[n];
for (int i = 0; i < n + 1; i++) {
a[i] = i;
}
for (int j = 2; j < n + 1; j++) {
if (a[j] != 0) {
for (int k = j * j; k < n + 1; k += j) {
a[k] = 0;
}
cout << a[j] << endl;
}
}
return 0;
}

Answer 1

You can find the biggest number in an array using std::max_element from <algorithm>:
return *std::max_element(a, a+n);
But this isn't even necessary. Since the biggest prime will be the last prime you found, just keep track of that:
int biggestPrime = 2;
And then, when you found a prime:
biggestPrime = j;
You can also replace your array of long long with an array of bool, since the index already keeps track of the number:
unsigned long long nPrime(unsigned n) {
bool *a = new bool[n + 1]();
long long maxNum = a[n];
int biggestPrime = 2;
for (int j = 2; j < n + 1; j++) {
if (!a[j]) {
for (int k = j * j; k < n + 1; k += j) {
a[k] = true;
}
biggestPrime = j;
}
}
delete[] a;
return biggestPrime;
}
And last but not least, don't forget to delete your dynamically allocated memory when you're done:
delete[] a;

Answer 2

As it seems you mark non-prime numbers with value 0 in the array. We know that primes are positive numbers, to simplify, let maxNum be -1. Then I added one for at the and of the function which will calculate the maximum prime value in the array. This is the one way to do :
unsigned long long nPrime(unsigned n) {
long long *a = new long long[n + 1];
long long maxNum = -1;
for (long long i = 0; i < n + 1; i++) {
a[i] = i;
}
for (long long i = 2; i < n + 1; i++) {
for (long long k = i * i; k < n + 1; k += i) {
a[k] = 0;
}
}
for (long long i = 1; i < n; i++) {
if (a[i] != 0 && maxNum < a[i]) maxNum = a[i];
}
return maxNum;
}
Also instead of last for you can use faster version. In this version you don't even need maxNum value : let's start from the end and first value which is non zero will be the answer:
for (long long i = n; i >= 2; i--) {
if (a[i] != 0) return a[i];
}