5.15~~Nice Clique

 https://www.hackerrank.com/blog/2020editorial

Given n numbers:

D = {d₁, d₂, …, d_n}

what’s the maximum size of a subset of D in which every pair is a ‘nice pair’?

(A, B) is a nice pair iff at least one of the following condition holds.
1. Parity of the number of prime divisors of A is equal to that of B.
2. Parity of the sum of all positive divisors of A is equal to that of B.

Input:
n
d₁ d₂ … d_n

Output:
The maximum size of the subset of D in which every pair is a nice pair

Constraints:
n ≤ 200
0 < d[i] ≤ 10^15

Sample Input:
4 
2 3 6 8

Sample Output:
3

Explanation:

d[i]	prime divisors (count)	divisors (sum)
2	2 (1)	1,2 (3)
3	3 (1)	1,3 (4)
6	2,3 (2)	1,2,3,6 (12)
8	2 (1)	1,2,4,8 (15)

==> (d₁, d₂) nice pair

(d₁, d₃) not

(d₁, d₄) nice

(d₂, d₃) nice

(d₂, d₄) nice

(d₃, d₄) not

==> max subset{2, 3, 8}. ==> S = 3.

 

Challenge 4: Nice Clique

We need to do exactly what described in the statement: find number of prime divisors and sum of all positive divisors.

Let's start with 80% solution and find prime divisors using "for" cycle from 2 to sqrt(n) and calculate sum of divisors using canonical number representation:

If x = p₁^α₁ * p₂^α₂ * ... * p_n^α_n, then sum of divisors of x is equal to

(1 + p₁ + p₁² + .. + p₁^α₁ * (1 + p₂ + ... + p₂^α₂) * ... * (1 + p_n + p_n² + ... + p_n^α_n) 

Proof: every summand is a divisor of x, and every divisor of x equals to some summand in this formula.

for (long j = 2; j * j <= a; j++)
{
    if (a % j == 0)
    {
        numberOfPrimeDivisors++;
        long currentValue = 1;
        long currentSum = 1;//(1 + x + x^2 + ..) * (1 + y + y ^2 + ...) + ...
        while (a % j == 0)
        {
            a /= j;
            currentValue *= j;
            currentSum += currentValue;
        }
        sumOfDivisors = (sumOfDivisors * currentSum) % 2;
    }
}

After we got these numbers, we can specify a class for each number:

1. c₀₀ -- (numberOfPrimeDivisors % 2 == 0) and (sumOfDivisors % 2 == 0)
2. c₀₁ -- (numberOfPrimeDivisors % 2 == 0) and (sumOfDivisors % 2 == 1)
3. c₁₀ -- (numberOfPrimeDivisors % 2 == 1) and (sumOfDivisors % 2 == 0)
4. c₁₁ -- (numberOfPrimeDivisors % 2 == 1) and (sumOfDivisors % 2 == 1)

Answer is the maximal sum of size of the two neigthboring(differing in one digit) classes(i.e. max(c₀₀ + c₀₁, c₀₀ + c₁₀, c₁₁ + c₁₀, c₁₁ + c₀₁), because if we take two random numbers from answer set:

 

1. If they belong to one class it means that both conditions are met
2. If they belong to "neighboring" class(c₀₀ and c₀₁, c₁₁ and c₁₀, but not, for example, c₀₀ and c₁₁) it means that one of these conditions is true
3. If they belong to (c₀₀ and c₁₁) or (c₀₁ and _c10) it means neither of these conditions is met and such elements coudn't be in answer set

 

To speed up this solution and get a full score I have generated list of prime divisors with Sieve of Eratosthenes/ up to sqrt(max number)

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

long long mul(long long x, long long y, long long mod) {
    if (x == 0) return 0; else {
        long long v = mul(x / 2, y, mod);
        return ((v+v)%mod+((x%2==0)?0:y))%mod;
    }
}
long long power_mod(long long x, long long e, long long mod) {
    return (e == 0) ? 1 : mul(power_mod(mul(x, x, mod), e / 2, mod), (e % 2 == 0) ? 1 : x, mod);
}

// Miller Rabin
bool isPrime(long long n) {
    int r = 0;
    long long R = n - 1;
    while (R % 2 == 0) {
        r++; R /= 2;
    }
    for (int it = 1; it <= 10; it++) {
        long long a = 1 + rand() % (n - 1);
        long long a_R (power_mod(a, R, n));
        vector<long long> b (r + 1);
        b[0] = a_R;
        for (int i = 1; i <= r; i++) {
            b[i] = mul(b[i - 1], b[i - 1], n);
        }
        if (b[r] != 1) {
            return false;
        }
        if (b[0] == 1) {
            continue;
        }
        int j = -1;
        for (int i = 0; i <= r; i++) {
            if (b[i] != 1) {
                j = i;
            }
        }
        if (b[j] == n - 1) {
            continue;
        }
        else {
            return false;
        }
    }
    return true;
}

bool isSquare(long long n) {
    long long l = 0, r = 1;
    while (r * r < n) r *= 2;
    while (l != r) {
        long long m = (l + r) / 2;
        (m * m < n) ? (l = m + 1) : (r = m);
    }
    return l * l == n;
}

int calc(long long n) {
    bool divisibleBy2 = n % 2 == 0;
    while (n % 2 == 0) n /= 2;
    // n odd
    int nFactors = int (divisibleBy2);
    int divSumParity = 1;
    for (int i = 2; (long long) i * i * i <= n; i ++) {
        if (n % i == 0) {
            int e = 0;
            while (n % i == 0) n /= i, e ++;
            if (e % 2 == 1) divSumParity = 0;
            nFactors ++;
        }
    }
    if (n > 1) {
        if (isSquare(n)) {
            nFactors += 1;
        } else if (isPrime(n)) {
            nFactors += 1;
            divSumParity = 0;
        } else {
            nFactors += 2;
            divSumParity = 0;
        }
    }
    return 2*(nFactors%2) + divSumParity;
}

inline int TWO(int n) { return 1 << n; }
inline int All1(int n) { return TWO(n) - 1; }

int main() {
    srand(0);
    vector<int> freq (4);
    int n;
    cin >> n;
    for (; n > 0; n --) {
        long long x;
        cin >> x;
        freq[calc(x)] ++;
    }
    cout << max(freq[0], freq[3]) + max(freq[1], freq[2]) << endl;
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */   
    return 0;
}