检查一个数字是否是一个完美的平方，其所有数字都是一个完美的平方

给定一个整数N ，任务是检查给定的数字是否为一个完美的平方，其所有数字均为完美的平方。如果发现是真的，则打印“是” 。否则，打印“否” 。
例子：

Input: N = 144 
Output: Yes 
Explanation: 
The number 144 is a perfect square and also the digits of the number {1(= 1², 4(= 2²} is also a perfect squares.
Input: N = 81 
Output: No

方法：想法是检查给定的数字N是否是一个完美的平方。如果发现为真，检查是否所有的位是0或者1，4，或9。如果发现是真的，则打印“是” 。否则，打印“否” 。
下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to check if digits of
// N is a perfect square or not
bool check_digits(long N)
{
 
    // Iterate over the digits
    while (N > 0) {
 
        // Extract the digit
        int n = N % 10;
 
        // Check if digit is a
        // perfect square or not
        if ((n != 0) && (n != 1)
            && (n != 4) && (n != 9)) {
            return 0;
        }
 
        // Divide N by 10
        N = N / 10;
    }
 
    // Return true
    return 1;
}
 
// Function to check if N is
// a perfect square or not
bool is_perfect(long N)
{
    long double n = sqrt(N);
 
    // If floor and ceil of n
    // is not same
    if (floor(n) != ceil(n)) {
        return 0;
    }
    return 1;
}
 
// Function to check if N satisfies
// the required conditions or not
void isFullSquare(long N)
{
    // If both the conditions
    // are satisfied
    if (is_perfect(N)
        && check_digits(N)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
}
 
// Driver Code
int main()
{
    long N = 144;
 
    // Function Call
    isFullSquare(N);
 
    return 0;
}

Java

// Java program for
// the above approach
import java.util.*;
class GFG{
 
// Function to check if digits of
// N is a perfect square or not
static boolean check_digits(long N)
{
  // Iterate over the digits
  while (N > 0)
  {
    // Extract the digit
    int n = (int) (N % 10);
 
    // Check if digit is a
    // perfect square or not
    if ((n != 0) && (n != 1) &&
        (n != 4) && (n != 9))
    {
      return false;
    }
 
    // Divide N by 10
    N = N / 10;
  }
 
  // Return true
  return true;
}
 
// Function to check if N is
// a perfect square or not
static boolean is_perfect(long N)
{
  double n = Math.sqrt(N);
 
  // If floor and ceil of n
  // is not same
  if (Math.floor(n) != Math.ceil(n))
  {
    return false;
  }
  return true;
}
 
// Function to check if N satisfies
// the required conditions or not
static void isFullSquare(long N)
{
  // If both the conditions
  // are satisfied
  if (is_perfect(N) &&
      check_digits(N))
  {
    System.out.print("Yes");
  }
  else
  {
    System.out.print("No");
  }
}
 
// Driver Code
public static void main(String[] args)
{
  long N = 144;
 
  // Function Call
  isFullSquare(N);
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 program for the above approach
import math
 
# Function to check if digits of
# N is a perfect square or not
def check_digits(N):
 
    # Iterate over the digits
    while (N > 0):
 
        # Extract the digit
        n = N % 10
 
        # Check if digit is a
        # perfect square or not
        if ((n != 0) and (n != 1) and
            (n != 4) and (n != 9)):
            return 0
     
        # Divide N by 10
        N = N // 10
     
    # Return true
    return 1
 
# Function to check if N is
# a perfect square or not
def is_perfect(N):
     
    n = math.sqrt(N)
 
    # If floor and ceil of n
    # is not same
    if (math.floor(n) != math.ceil(n)):
        return 0
     
    return 1
 
# Function to check if N satisfies
# the required conditions or not
def isFullSquare(N):
     
    # If both the conditions
    # are satisfied
    if (is_perfect(N) and
      check_digits(N)):
        print("Yes")
    else:
        print("No")
     
# Driver Code
N = 144
 
# Function call
isFullSquare(N)
 
# This code is contributed by sanjoy_62

C#

// C# program for
// the above approach
using System;
class GFG{
 
// Function to check if digits of
// N is a perfect square or not
static bool check_digits(long N)
{
  // Iterate over the digits
  while (N > 0)
  {
    // Extract the digit
    int n = (int) (N % 10);
 
    // Check if digit is a
    // perfect square or not
    if ((n != 0) && (n != 1) &&
        (n != 4) && (n != 9))
    {
      return false;
    }
 
    // Divide N by 10
    N = N / 10;
  }
 
  // Return true
  return true;
}
 
// Function to check if N is
// a perfect square or not
static bool is_perfect(long N)
{
  double n = Math.Sqrt(N);
 
  // If floor and ceil of n
  // is not same
  if (Math.Floor(n) != Math.Ceiling(n))
  {
    return false;
  }
  return true;
}
 
// Function to check if N satisfies
// the required conditions or not
static void isFullSquare(long N)
{
  // If both the conditions
  // are satisfied
  if (is_perfect(N) &&
      check_digits(N))
  {
    Console.Write("Yes");
  }
  else
  {
    Console.Write("No");
  }
}
 
// Driver Code
public static void Main()
{
  long N = 144;
 
  // Function Call
  isFullSquare(N);
}
}
 
// This code is contributed by Chitranayal

Javascript

 
// JavaScript program for the above approach
 
// Function to check if digits of
// N is a perfect square or not
function check_digits(N)
{
 
    // Iterate over the digits
    while (N > 0) {
 
        // Extract the digit
        let n = N % 10;
 
        // Check if digit is a
        // perfect square or not
        if ((n != 0) && (n != 1)
            && (n != 4) && (n != 9)) {
            return 0;
        }
 
        // Divide N by 10
        N = Math.floor(N / 10);
    }
 
    // Return true
    return 1;
}
 
// Function to check if N is
// a perfect square or not
function is_perfect(N)
{
    let n = Math.sqrt(N);
 
    // If floor and ceil of n
    // is not same
    if (Math.floor(n) != Math.ceil(n)) {
        return 0;
    }
    return 1;
}
 
// Function to check if N satisfies
// the required conditions or not
function isFullSquare(N)
{
    // If both the conditions
    // are satisfied
    if (is_perfect(N)
        && check_digits(N)) {
        document.write("Yes");
    }
    else {
        document.write("No");
    }
}
 
// Driver Code
 
    let N = 144;
 
    // Function Call
    isFullSquare(N);
 
 
// This code is contributed by Surbhi Tyagi.
 

输出：

Yes

时间复杂度： O(log ₁₀ N)
辅助空间： O(1)