Codility-CommonPrimeDivisors

Task description

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..2147483647].

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).

Elements of input arrays can be modified.

思路：a和b共有质因子必然全都是a和b最大公约数gcd（a，b）的因子。若a和b有完全相同的质因子集，则a和b所有的质因子都是gcd（a，b）的质因子，让a不断除以gcd（a，b）的每一个质因子后（如b=75，gcd（a，b）=15，gcd（a，b）的质因子为3和5，b=b/3=25，b=b/5=5,b=b/5=1），最终会得到1，对b也是如此。故最终若a=b=1，则a和b有相同的质因子集，否则没有。

代码：