给定一个由N个非负整数组成的数组arr [] ，任务是计算其数字乘积为整数的数组元素的数量。
例子：
Input: arr[] = {13, 55, 7, 13, 11, 71, 233, 144, 89}
Output: 4
Explanation: The array elements having product of digits equal to a composite number are 55 (5 * 5 = 25), 233 (2 * 3 * 3 = 18), 144 (1 * 4 * 4 = 16), 89 ( = 8 * 9 = 72)
Input: arr[] = {34, 13, 55, 11, 8, 3, 55, 23}
Output: 3
Explanation: The array elements having product of digits equal to a composite number are 34 ( = 3 * 4 = 12), 55 ( 5 * 5 = 25), 23 ( = 2 * 3 = 6)
方法：请按照以下步骤解决问题：
- 使用Eratosthenes筛子预先计算并存储所有质数。
- 创建一个Set来存储所有不同的数组元素，这些元素的数字乘积是一个复合数字。
- 遍历给定数组，对于每个数组元素，检查其所有数字的乘积是否为质数。如果发现为真，则将当前元素添加到Set中。
- 否则，继续下一个数组元素。
- 遍历后，将设置的大小打印为所需的答案。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
#include
using namespace std;
#define N 100005
// Function to generate prime numbers
// using Sieve of Eratosthenes
void SieveOfEratosthenes(bool prime[],
int p_size)
{
// Set 0 and 1 as non-prime
prime[0] = false;
prime[1] = false;
for (int p = 2; p * p <= p_size; p++) {
// If p is a prime
if (prime[p]) {
// Set all multiples of p as non-prime
for (int i = p * 2; i <= p_size; i += p)
prime[i] = false;
}
}
}
// Function to calculate the product
// of digits of the given number
long long int digitProduct(int number)
{
// Stores the product of digits
long long int res = 1;
while (number > 0) {
// Extract digits and
// add to the sum
res *= (number % 10);
number /= 10;
}
// Return the product
// of its digits
return res;
}
// Function to print number of distinct
// values with digit product as composite
void DistinctCompositeDigitProduct(int arr[],
int n)
{
// Initialize set
set output;
// Initialize boolean array
bool prime[N + 1];
memset(prime, true, sizeof(prime));
// Pre-compute primes
SieveOfEratosthenes(prime, N);
// Traverse array
for (int i = 0; i < n; i++) {
// Stores the product of digits
// of the current array element
long long int ans
= digitProduct(arr[i]);
// If Product of digits is
// less than or equal to 1
if (ans <= 1) {
continue;
}
// If Product of digits is
// not a prime
if (!prime[ans]) {
output.insert(ans);
}
}
// Print the answer
cout << output.size() << endl;
}
// Driver Code
int main()
{
// Given array
int arr[]
= { 13, 55, 7, 13, 11,
71, 233, 233, 144, 89 };
// Given size
int n = sizeof(arr)
/ sizeof(arr[0]);
// Function call
DistinctCompositeDigitProduct(arr, n);
return 0;
} Java
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG{
static int N = 100005;
// Function to generate prime numbers
// using Sieve of Eratosthenes
static void SieveOfEratosthenes(boolean prime[],
int p_size)
{
// Set 0 and 1 as non-prime
prime[0] = false;
prime[1] = false;
for(int p = 2; p * p <= p_size; p++)
{
// If p is a prime
if (prime[p])
{
// Set all multiples of p as non-prime
for(int i = p * 2; i <= p_size; i += p)
prime[i] = false;
}
}
}
// Function to calculate the product
// of digits of the given number
static int digitProduct(int number)
{
// Stores the product of digits
int res = 1;
while (number > 0)
{
// Extract digits and
// add to the sum
res *= (number % 10);
number /= 10;
}
// Return the product
// of its digits
return res;
}
// Function to print number of distinct
// values with digit product as composite
static void DistinctCompositeDigitProduct(int arr[],
int n)
{
// Initialize set
TreeSet output = new TreeSet();
// Initialize boolean array
boolean prime[] = new boolean[N + 1];
Arrays.fill(prime, true);
// Pre-compute primes
SieveOfEratosthenes(prime, N);
// Traverse array
for(int i = 0; i < n; i++)
{
// Stores the product of digits
// of the current array element
int ans = digitProduct(arr[i]);
// If Product of digits is
// less than or equal to 1
if (ans <= 1)
{
continue;
}
// If Product of digits is
// not a prime
if (!prime[ans])
{
output.add(ans);
}
}
// Print the answer
System.out.print(output.size());
}
// Driver Code
public static void main(String[] args)
{
// Given array
int arr[] = { 13, 55, 7, 13, 11,
71, 233, 233, 144, 89 };
// Given size
int n = arr.length;
// Function call
DistinctCompositeDigitProduct(arr, n);
}
}
// This code is contributed by susmitakundugoaldanga Python3
# Python3 program for the above approach
N = 100005
from math import sqrt
# Function to generate prime numbers
# using Sieve of Eratosthenes
def SieveOfEratosthenes(prime, p_size):
# Set 0 and 1 as non-prime
prime[0] = False
prime[1] = False
for p in range(2, int(sqrt(p_size)), 1):
# If p is a prime
if (prime[p]):
# Set all multiples of p as non-prime
for i in range(p * 2, p_size + 1, p):
prime[i] = False
# Function to calculate the product
# of digits of the given number
def digitProduct(number):
# Stores the product of digits
res = 1
while(number > 0):
# Extract digits and
# add to the sum
res *= (number % 10)
number //= 10
# Return the product
# of its digits
return res
# Function to print number of distinct
# values with digit product as composite
def DistinctCompositeDigitProduct(arr, n):
# Initialize set
output = set()
# Initialize boolean array
prime = [True for i in range(N + 1)]
# Pre-compute primes
SieveOfEratosthenes(prime, N)
# Traverse array
for i in range(n):
# of the current array element
ans = digitProduct(arr[i])
# If Product of digits is
# less than or equal to 1
if (ans <= 1):
continue
# If Product of digits is
# not a prime
if (prime[ans] == False):
output.add(ans)
# Print the answer
print(len(output))
# Driver Code
if __name__ == '__main__':
# Given array
arr = [13, 55, 7, 13, 11, 71, 233, 233, 144, 89]
# Given size
n = len(arr)
# Function call
DistinctCompositeDigitProduct(arr, n)
# This code is contributed by SURENDRA_GANGWAR.C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
static int N = 100005;
// Function to generate prime numbers
// using Sieve of Eratosthenes
static void SieveOfEratosthenes(bool[] prime,
int p_size)
{
// Set 0 and 1 as non-prime
prime[0] = false;
prime[1] = false;
for(int p = 2; p * p <= p_size; p++)
{
// If p is a prime
if (prime[p])
{
// Set all multiples of p as non-prime
for(int i = p * 2; i <= p_size; i += p)
prime[i] = false;
}
}
}
// Function to calculate the product
// of digits of the given number
static int digitProduct(int number)
{
// Stores the product of digits
int res = 1;
while (number > 0)
{
// Extract digits and
// add to the sum
res *= (number % 10);
number /= 10;
}
// Return the product
// of its digits
return res;
}
// Function to print number of distinct
// values with digit product as composite
static void DistinctCompositeDigitProduct(int[] arr,
int n)
{
// Initialize set
SortedSet output = new SortedSet();
// Initialize boolean array
bool[] prime = new bool[N + 1];
for(int i = 0; i < N + 1; i++)
{
prime[i] = true;
}
// Pre-compute primes
SieveOfEratosthenes(prime, N);
// Traverse array
for(int i = 0; i < n; i++)
{
// Stores the product of digits
// of the current array element
int ans = digitProduct(arr[i]);
// If Product of digits is
// less than or equal to 1
if (ans <= 1)
{
continue;
}
// If Product of digits is
// not a prime
if (!prime[ans])
{
output.Add(ans);
}
}
// Print the answer
Console.WriteLine(output.Count);
}
// Driver Code
public static void Main()
{
// Given array
int[] arr = { 13, 55, 7, 13, 11,
71, 233, 233, 144, 89 };
// Given size
int n = arr.Length;
// Function call
DistinctCompositeDigitProduct(arr, n);
}
}
// This code is contributed by sanjoy_62 输出：
4时间复杂度： O(N + MlogM)，其中N是给定数组的大小， M是最大数组元素。
辅助空间： O(1)