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:
- 初始化变量sum,以存储数组元素的位数的总和。
- 遍历数组并将每个数组元素转换为其等效的字符串
- 将每个字符串的长度相加。
- 检查数组完全遍历后的sum值是否为质数。
- 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)