查找其乘积具有最大素因子数的行

给定一个大小为N x M的矩阵，任务是打印其乘积具有最大素因子数的行的元素。
例子：

Input: arr[][] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}; 
Output: 7 8 9 
Explanation: 
Row 1: (1, 2, 3) has product 6 and it has 2 prime factors. 
Row 2: (4, 5, 6) has product 120 and it has 3 prime factors. 
Row 3: (7, 8, 9) has product 504 and it has 6 prime factors. 
Therefore, the output is 7 8 9, as it has maximum count of prime factors.
Input: arr[][] = {{11, 12, 13}, {14, 15, 16}, {17, 18, 19}} 
Output: 14 15 16 
 

方法：

• 通过遍历所有元素并找到它们的质因数，找出每行中每个质因数的总出现次数。我们使用散列来计算出现次数。
• 设质因数出现的次数为a1, a2, …aK 。如果我们有 K 个不同的质因数，那么答案将是：

  $(a_{1}+1)(a_{2}+1)(...)*(a_{K}+1)$

• 将此值与max_factor中连续存储最大素因子数的值进行比较。如果更大，则更新该行的值。
• 继续直到遍历所有行。
  下面是上述方法的实现：

C++

// C++ implementation to find the row
// whose product has maximum number
// of prime factors
  
#include 
using namespace std;
  
#define N 3
#define M 5
  
int Large = 1e6;
  
vector prime;
  
// function for SieveOfEratosthenes
void SieveOfEratosthenes()
{
  
    // Create a boolean array "isPrime[0..N]"
    // and initialize all entries it as true.
    // A value in isPrime[i] will finally be
    // false if i is not a prime, else true.
    bool isPrime[Large + 1];
    memset(isPrime, true, sizeof(isPrime));
  
    for (int p = 2; p * p <= Large; p++) {
  
        // check if isPrime[p] is not changed
        if (isPrime[p] == true) {
  
            // Update all multiples of p
            for (int i = p * 2; i <= Large; i += p)
                isPrime[i] = false;
        }
    }
  
    // Print all isPrime numbers
    for (int p = 2; p <= Large; p++)
  
        if (isPrime[p])
  
            prime.push_back(p);
}
  
// function to display the answer
void Display(int arr[][M], int row)
{
  
    for (int i = 0; i < M; i++)
        cout << arr[row][i] << " ";
}
  
// function to Count the row number of
// divisors in particular row multiplication
void countDivisorsMult(int arr[][M])
{
  
    // Find count of occurrences
    // of each prime factor
    unordered_map mp;
    int row_no = 0;
    long long max_factor = 0;
  
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < M; j++) {
            int no = arr[i][j];
  
            for (int k = 0; k < prime.size(); k++) {
                while (no > 1
                       && no % prime[k] == 0) {
  
                    no /= prime[k];
                    mp[prime[k]]++;
                }
  
                if (no == 1)
                    break;
            }
        }
  
        // Compute count of all divisors
        long long int res = 1;
        for (auto it : mp) {
            res *= (it.second + 1L);
        }
  
        // Update row number if
        // factors of this row is max
        if (max_factor < res) {
            row_no = i;
            max_factor = res;
        }
  
        // Clearing map to store
        // prime factors for next row
        mp.clear();
    }
  
    Display(arr, row_no);
}
  
// Driver code
int main()
{
  
    int arr[N][M] = { { 1, 2, 3, 10, 23 },
                      { 4, 5, 6, 7, 8 },
                      { 7, 8, 9, 15, 45 } };
  
    SieveOfEratosthenes();
  
    countDivisorsMult(arr);
  
    return 0;
}

Java

// Java implementation to find the row
// whose product has maximum number
// of prime factors
import java.util.*;
  
class GFG{
   
static final int N = 3;
static final int M = 5;
   
static int Large = (int) 1e6;
   
static Vector prime = new Vector();
   
// function for SieveOfEratosthenes
static void SieveOfEratosthenes()
{
   
    // Create a boolean array "isPrime[0..N]"
    // and initialize all entries it as true.
    // A value in isPrime[i] will finally be
    // false if i is not a prime, else true.
    boolean []isPrime = new boolean[Large + 1];
    Arrays.fill(isPrime, true);
   
    for (int p = 2; p * p <= Large; p++) {
   
        // check if isPrime[p] is not changed
        if (isPrime[p] == true) {
   
            // Update all multiples of p
            for (int i = p * 2; i <= Large; i += p)
                isPrime[i] = false;
        }
    }
   
    // Print all isPrime numbers
    for (int p = 2; p <= Large; p++)
   
        if (isPrime[p])
   
            prime.add(p);
}
   
// function to display the answer
static void Display(int arr[][], int row)
{
   
    for (int i = 0; i < M; i++)
        System.out.print(arr[row][i]+ " ");
}
   
// function to Count the row number of
// divisors in particular row multiplication
static void countDivisorsMult(int arr[][])
{
   
    // Find count of occurrences
    // of each prime factor
    HashMap mp = new HashMap();
    int row_no = 0;
    long max_factor = 0;
   
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < M; j++) {
            int no = arr[i][j];
   
            for (int k = 0; k < prime.size(); k++) {
                while (no > 1
                       && no % prime.get(k) == 0) {
   
                    no /= prime.get(k);
                    if(mp.containsKey(prime.get(k)))
                        mp.put(prime.get(k), prime.get(k)+1);
                    else
                        mp.put(prime.get(k), 1);
                }
   
                if (no == 1)
                    break;
            }
        }
   
        // Compute count of all divisors
        int res = 1;
        for (Map.Entry it : mp.entrySet()) {
            res *= (it.getValue() + 1L);
        }
   
        // Update row number if
        // factors of this row is max
        if (max_factor < res) {
            row_no = i;
            max_factor = res;
        }
   
        // Clearing map to store
        // prime factors for next row
        mp.clear();
    }
   
    Display(arr, row_no);
}
   
// Driver code
public static void main(String[] args)
{
   
    int arr[][] = { { 1, 2, 3, 10, 23 },
                      { 4, 5, 6, 7, 8 },
                      { 7, 8, 9, 15, 45 } };
   
    SieveOfEratosthenes();
   
    countDivisorsMult(arr);
   
}
}
  
// This code is contributed by Rajput-Ji

Python3

# Python3 implementation to find the row 
# whose product has maximum number 
# of prime factors 
N = 3
M = 5
  
Large = int(1e6); 
  
prime = []; 
  
# function for SieveOfEratosthenes 
def SieveOfEratosthenes() :
  
    # Create a boolean array "isPrime[0..N]" 
    # and initialize all entries it as true. 
    # A value in isPrime[i] will finally be 
    # false if i is not a prime, else true. 
    isPrime = [True]*(Large + 1); 
      
    for p in range(2, int(Large**(1/2))) : 
  
        # check if isPrime[p] is not changed 
        if (isPrime[p] == True) :
  
            # Update all multiples of p 
            for i in range(p*2, Large + 1, p) : 
                isPrime[i] = False; 
  
    # Print all isPrime numbers 
    for p in range(2, Large + 1) :
  
        if (isPrime[p]) :
  
            prime.append(p); 
  
# function to display the answer 
def Display(arr, row) : 
  
    for i in range(M) : 
        print(arr[row][i], end=" "); 
  
# function to Count the row number of 
# divisors in particular row multiplication 
def countDivisorsMult(arr) : 
  
    # Find count of occurrences 
    # of each prime factor 
    mp = {};
    row_no = 0;max_factor = 0; 
  
    for i in range(N) :
        for j in range(M) : 
            no = arr[i][j]
              
            for k in range(len(prime)) :
                while (no > 1 and no % prime[k] == 0) :
                      
                    no //= prime[k];
                      
                    if prime[k] not in mp :
                        mp[prime[k]] = 0
                      
                    mp[prime[k]] += 1;
                      
                if (no == 1) :
                    break; 
  
        # Compute count of all divisors 
        res = 1; 
        for it in mp :
            res *= mp[it]; 
  
        # Update row number if 
        # factors of this row is max 
        if (max_factor < res) :
            row_no = i; 
            max_factor = res; 
          
        # Clearing map to store 
        # prime factors for next row 
        mp.clear(); 
  
    Display(arr, row_no); 
  
# Driver code 
if __name__ == "__main__" : 
  
  
    arr = [ [ 1, 2, 3, 10, 23 ], 
            [ 4, 5, 6, 7, 8 ], 
            [ 7, 8, 9, 15, 45 ] ]; 
  
    SieveOfEratosthenes(); 
  
    countDivisorsMult(arr); 
  
# This code is contributed by Yash_R

C#

// C# implementation to find the row
// whose product has maximum number
// of prime factors
using System;
using System.Collections.Generic;
class GFG{ 
static readonly int N = 3;
static readonly int M = 5; 
static int Large = (int) 1e6; 
static List prime = new List();
   
// function for SieveOfEratosthenes
static void SieveOfEratosthenes()
{ 
    // Create a bool array "isPrime[0..N]"
    // and initialize all entries it as true.
    // A value in isPrime[i] will finally be
    // false if i is not a prime, else true.
    bool []isPrime = new bool[Large + 1];
    for (int p = 0; p <= Large; p++)
        isPrime[p] = true;
   
    for (int p = 2; p * p <= Large; p++) 
    { 
        // check if isPrime[p] is not changed
        if (isPrime[p] == true) 
        { 
            // Update all multiples of p
            for (int i = p * 2; i <= Large; i += p)
                isPrime[i] = false;
        }
    }
   
    // Print all isPrime numbers
    for (int p = 2; p <= Large; p++) 
        if (isPrime[p]) 
            prime.Add(p);
}
   
// function to display the answer
static void Display(int [, ]arr, int row)
{ 
    for (int i = 0; i < M; i++)
        Console.Write(arr[row, i] + " ");
}
   
// function to Count the row number of
// divisors in particular row multiplication
static void countDivisorsMult(int [, ]arr)
{ 
    // Find count of occurrences
    // of each prime factor
    Dictionary mp = new Dictionary();
    int row_no = 0;
    long max_factor = 0; 
    for (int i = 0; i < N; i++) 
    {
        for (int j = 0; j < M; j++) 
        {
            int no = arr[i,j]; 
            for (int k = 0; k < prime.Count; k++) 
            {
                while (no > 1 && no % 
                       prime[k] == 0) 
                { 
                    no /= prime[k];
                    if(mp.ContainsKey(prime[k]))
                        mp[prime[k]] = prime[k] + 1;
                    else
                        mp.Add(prime[k], 1);
                }
   
                if (no == 1)
                    break;
            }
        }
   
        // Compute count of all divisors
        int res = 1;
        foreach (KeyValuePair it in mp) 
        {
            res *= (it.Value + 1);
        }
   
        // Update row number if
        // factors of this row is max
        if (max_factor < res) 
        {
            row_no = i;
            max_factor = res;
        }
   
        // Clearing map to store
        // prime factors for next row
        mp.Clear();
    } 
    Display(arr, row_no);
}
   
// Driver code
public static void Main(String[] args)
{ 
    int [, ]arr = {{1, 2, 3, 10, 23},
                  {4, 5, 6, 7, 8},
                  {7, 8, 9, 15, 45}}; 
    SieveOfEratosthenes(); 
    countDivisorsMult(arr);
}
}
  
// This code is contributed by Rajput-Ji

Javascript

  
// JavaScript implementation to find the row
// whose product has maximum number
// of prime factors
  
  
let N  = 3
let M = 5
  
let Large = 1e6;
  
let prime = new Array();
  
// function for SieveOfEratosthenes
function SieveOfEratosthenes()
{
  
    // Create a boolean array "isPrime[0..N]"
    // and initialize all entries it as true.
    // A value in isPrime[i] will finally be
    // false if i is not a prime, else true.
    let isPrime = new Array();
  
    for(let i = 0;  i < Large + 1; i++){
        isPrime.push([])
    }
  
    isPrime.fill(true);
  
    for (let p = 2; p * p <= Large; p++) {
  
        // check if isPrime[p] is not changed
        if (isPrime[p] == true) {
  
            // Update all multiples of p
            for (let i = p * 2; i <= Large; i += p)
                isPrime[i] = false;
        }
    }
  
    // Print all isPrime numbers
    for (let p = 2; p <= Large; p++)
  
        if (isPrime[p])
  
            prime.push(p);
}
  
// function to display the answer
function Display(arr, row)
{
  
    for (let i = 0; i < M; i++)
        document.write(arr[row][i] + " ");
}
  
// function to Count the row number of
// divisors in particular row multiplication
function countDivisorsMult(arr)
{
  
    // Find count of occurrences
    // of each prime factor
    let mp = new Map();
    let row_no = 0;
    let max_factor = 0;
  
    for (let i = 0; i < N; i++) {
        for (let j = 0; j < M; j++) {
            let no = arr[i][j];
  
            for (let k = 0; k < prime.length; k++) {
                while (no > 1 && no % prime[k] == 0) {
  
                    no /= prime[k];
                    if(mp.has(prime[k])){
                        mp.set(prime[k], mp.get(prime[k]) + 1)
                    }else{
                        mp.set(prime[k], 1)
                    }
                }
  
                if (no == 1)
                    break;
            }
        }
  
        // Compute count of all divisors
        let res = 1;
        for (let it of mp) {
            res *= (it[1] + 1);
        }
  
        // Update row number if
        // factors of this row is max
        if (max_factor < res) {
            row_no = i;
            max_factor = res;
        }
  
        // Clearing map to store
        // prime factors for next row
        mp.clear();
    }
  
    Display(arr, row_no);
}
  
// Driver code
  
  
let arr = [ [ 1, 2, 3, 10, 23 ],
            [ 4, 5, 6, 7, 8 ],
            [ 7, 8, 9, 15, 45 ] ];
  
SieveOfEratosthenes();
  
countDivisorsMult(arr);
  
// This code is contributed by gfgking
  

输出：

7 8 9 15 45

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