LintCode 1295: Prime Factor Statistics (质因数分解经典题)

1295. Prime Factor Statistics

 

Given a positive integer N, you need to factorize all integers between (1, N].

Then you have to count the number of the total prime numbers.

 

Example

input：6

output：7

explain：2=2, 3=3, 4=2*2, 5=5, 6=2*3, the number of prime number : 1+1+2+1+2=7

Notice

1<N<=100000

解法1：DP。总时间复杂度是O(n * sqrt(n))。

class Solution {
public:
    /**
     * @param N: a number
     * @return: the number of prime numbers.
     */
    int Count_PrimeNum(int N) {
        int result = 0;
        vector<int> prime(N + 1, 1); prime[0] = 0; prime[1] = 0;
        vector<bool> visited(N + 1, false);
        for (int i = 2; i <= N; ++i) {
            result += prime[i];
            for (int j = 2; i * j <= N && j <= i; ++j) {
                if (!visited[i * j]) {
                    visited[i * j] = 1;
                    prime[i * j] = prime[i] + prime[j];
                }
            }       
        }
        
        return result;
    }
};

二刷，不需要visited数组。

class Solution {
public:
    /**
     * @param N: a number
     * @return: the number of prime numbers.
     */
    int Count_PrimeNum(int N) {
        int result = 0;
        vector<int> records(N + 1, 1);
        records[0] = 0, records[1] = 0;
        for (int i = 2; i <= N; ++i) {
            result += records[i];
                int count = N / i;
                for (int j = 1; j <= count; ++j) {
                    records[i * j] = records[i] + records[j];
                }
        }
        
        return result;
    }
};

 

解法2：基于Pollard_Rho算法。因为Pollard_Rho算法时间复杂度O(n^ 1/4)，所以总的时间复杂度是O(n^ 5/4)。

class Solution {
public:
    /**
     * @param N: a number
     * @return: the number of prime numbers.
     */
    int Count_PrimeNum(int N) {
        int result = 0;
        vector<int> nums(N + 1, -1);
    
        for (int i = 2; i <= N; ++i) {
            result += Pollard_Rho(i);
        } 
        return result;
    }
    
private:
    int Pollard_Rho(int n) {
        int count = 0;
        //vector<int> result;
        int sqrtn = sqrt(n);
        for (int i = 2; i <= sqrtn;) {
            while (n % i == 0) {
                count++;
             //   result.push_back(i);
                n /= i;
            }
            
            if (i == 2) i += 1;
            else i += 2;
        }
      //  if (count == 0) count = 1;  // if n is non-2 prime
        if (n > 1) count++;
        return count;
    }
};