[Codility] CommonPrimeDivisors

A prime is a positive integer X that has exactly two distinct divisors: 1 and X. The first few prime integers are 2, 3, 5, 7, 11 and 13.

A prime D is called a prime divisor of a positive integer P if there exists a positive integer K such that D * K = P. For example, 2 and 5 are prime divisors of 20.

You are given two positive integers N and M. The goal is to check whether the sets of prime divisors of integers N and M are exactly the same.

For example, given:

• N = 15 and M = 75, the prime divisors are the same: {3, 5};
• N = 10 and M = 30, the prime divisors aren't the same: {2, 5} is not equal to {2, 3, 5};
• N = 9 and M = 5, the prime divisors aren't the same: {3} is not equal to {5}.

Write a function:

int solution(vector<int> &A, vector<int> &B);

that, given two non-empty zero-indexed arrays A and B of Z integers, returns the number of positions K for which the prime divisors of A[K] and B[K] are exactly the same.

For example, given:

    A[0] = 15   B[0] = 75
    A[1] = 10   B[1] = 30
    A[2] = 3    B[2] = 5

the function should return 1, because only one pair (15, 75) has the same set of prime divisors.

Assume that:

• Z is an integer within the range [1..6,000];
• each element of arrays A, B is an integer within the range [1..2,147,483,647].

Complexity:

• expected worst-case time complexity is O(Z*log(max(A)+max(B))2);
• expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments).

判断两个数是否有相同的素数约数。首先求出公约数gcd_val，那么gcd_val里应该包含了common prime divisor，下面分别判断a跟b与gcd_val的公约数是不是有自己的非common prime divisor的prime divisor。

// you can use includes, for example:
// #include <algorithm>

// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;
int gcd(int a, int b) {
    if (a < b) return gcd(b, a);
    return b > 0 ? gcd(b, a % b) : a;
}

bool hasSamePrimeDivisors(int a, int b) {
    int gcd_val = gcd(a, b);
    int gcd_a, gcd_b;
    while (a != 1) {
        gcd_a = gcd(a, gcd_val);
        if (gcd_a == 1) break;
        a /= gcd_a;
    }
    if (a != 1) return false;
    while (b != 1) {
        gcd_b = gcd(b, gcd_val);
        if (gcd_b == 1) break;
        b /= gcd_b;
    }
    return b == 1;
}

int solution(vector<int> &A, vector<int> &B) {
    // write your code in C++11
    int cnt = 0;
    for (int i = 0; i < A.size() && i < B.size(); ++i) {
        if (hasSamePrimeDivisors(A[i], B[i])) ++cnt;
    }
    return cnt;
}

def gcd(x, y):
    # Compute the greatest common divisor
    if x%y == 0:
        return y;
    else:
        return gcd(y, x%y)

def hasSamePrimeDivisors(x, y):
    gcd_value = gcd(x, y)   # The gcd contains all
                            # the common prime divisors

    while x != 1:
        x_gcd = gcd(x, gcd_value)
        if x_gcd == 1:
            # x does not contain any more 
            # common prime divisors
            break
        x /= x_gcd
    if x != 1:
        # If x and y have exactly the same common 
        # prime divisors, x must be composed by
        # the prime divisors in gcd_value. So
        # after previous loop, x must be one.
        return False

    while y != 1:
        y_gcd = gcd(y, gcd_value)
        if y_gcd == 1:
            # y does not contain any more 
            # common prime divisors
            break
        y /= y_gcd

    return y == 1

def solution(A, B):
    count = 0
    for x,y in zip(A,B):
        if hasSamePrimeDivisors(x,y):
            count += 1
    return count