检查数组元素的位数总和是否为质数

给定一个由N个整数组成的数组A [] ，任务是检查每个数组元素中的位数之和是否为质数。

例子：

Input: A[] = {1, 11, 12}
Output: Yes
Explanation: Count of digits of A[0], A[1] and A[2] are 1, 2, 2 respectively. Therefore, total sum of count of digits = 1 + 2 + 2 = 5, which is prime.

Input: A[] = {1, 11, 123}
Output: No

方法：请按照以下步骤解决问题：

1. 初始化变量sum，以存储数组元素的位数的总和。
2. 遍历数组并将每个数组元素转换为其等效的字符串
3. 将每个字符串的长度相加。
4. 检查数组完全遍历后的sum值是否为质数。
5. 如果发现是真的，则打印“是” 。否则，打印No。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to check whether
// a number is prime or not
bool isPrime(int n)
{
    // Corner cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;
 
    // If given number is a
    // multiple of 2 or 3
    if (n % 2 == 0 || n % 3 == 0)
        return false;
 
    for (int i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;
 
    return true;
}
 
// Function to check if sum
// of count of digits of all
// array elements is prime or not
void CheckSumPrime(int A[], int N)
{
    // Initialize sum with 0
    int sum = 0;
 
    // Traverse over the array
    for (int i = 0; i < N; i++) {
 
        // Convert array element to string
        string s = to_string(A[i]);
 
        // Add the count of
        // digits to sum
        sum += s.length();
    }
 
    // Print the result
    if (isPrime(sum)) {
 
        cout << "Yes" << endl;
    }
    else {
 
        cout << "No" << endl;
    }
}
 
// Drive Code
int main()
{
 
    int A[] = { 1, 11, 12 };
 
    int N = sizeof(A) / sizeof(A[0]);
 
    // Function call
    CheckSumPrime(A, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG{
 
// Function to check whether
// a number is prime or not
static boolean isPrime(int n)
{
     
    // Corner cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;
 
    // If given number is a
    // multiple of 2 or 3
    if (n % 2 == 0 || n % 3 == 0)
        return false;
 
    for(int i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;
 
    return true;
}
 
// Function to check if sum
// of count of digits of all
// array elements is prime or not
static void CheckSumPrime(int[] A, int N)
{
     
    // Initialize sum with 0
    int sum = 0;
 
    // Traverse over the array
    for(int i = 0; i < N; i++)
    {
         
        // Convert array element to string
        String s = Integer.toString(A[i]);
 
        // Add the count of
        // digits to sum
        sum += s.length();
    }
 
    // Print the result
    if (isPrime(sum) == true)
    {
        System.out.println("Yes");
    }
    else
    {
        System.out.println("No");
    }
}
 
// Drive Code
public static void main(String[] args)
{
    int[] A = { 1, 11, 12 };
 
    int N = A.length;
 
    // Function call
    CheckSumPrime(A, N);
}
}
 
// This code is contributed by akhilsaini

Python3

# Python3 program for the above approach
import math
 
# Function to check whether
# a number is prime or not
def isPrime(n):
     
    # Corner cases
    if (n <= 1):
        return False
    if (n <= 3):
        return True
 
    # If given number is a
    # multiple of 2 or 3
    if (n % 2 == 0 or n % 3 == 0):
        return False
 
    for i in range(5, int(math.sqrt(n) + 1), 6):
        if (n % i == 0 or n % (i + 2) == 0):
            return False
 
    return True
 
# Function to check if sum
# of count of digits of all
# array elements is prime or not
def CheckSumPrime(A, N):
     
    # Initialize sum with 0
    sum = 0
 
    # Traverse over the array
    for i in range(0, N):
         
        # Convert array element to string
        s = str(A[i])
 
        # Add the count of
        # digits to sum
        sum += len(s)
 
    # Print the result
    if (isPrime(sum) == True):
        print("Yes")
    else:
        print("No")
 
# Drive Code
if __name__ == '__main__':
     
    A = [ 1, 11, 12 ]
 
    N = len(A)
 
    # Function call
    CheckSumPrime(A, N)
 
# This code is contributed by akhilsaini

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to check whether
// a number is prime or not
static bool isPrime(int n)
{
     
    // Corner cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;
 
    // If given number is a
    // multiple of 2 or 3
    if (n % 2 == 0 || n % 3 == 0)
        return false;
 
    for(int i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;
 
    return true;
}
 
// Function to check if sum
// of count of digits of all
// array elements is prime or not
static void CheckSumPrime(int[] A, int N)
{
     
    // Initialize sum with 0
    int sum = 0;
 
    // Traverse over the array
    for(int i = 0; i < N; i++)
    {
         
        // Convert array element to string
        String s = A[i].ToString();
 
        // Add the count of
        // digits to sum
        sum += s.Length;
    }
 
    // Print the result
    if (isPrime(sum) == true)
    {
        Console.WriteLine("Yes");
    }
    else
    {
        Console.WriteLine("No");
    }
}
 
// Drive Code
public static void Main()
{
    int[] A = { 1, 11, 12 };
 
    int N = A.Length;
 
    // Function call
    CheckSumPrime(A, N);
}
}
 
// This code is contributed by akhilsaini

输出：

Yes

时间复杂度： O(N ^3/2 )
辅助空间： O(1)