给定自然数N ，任务是找到最大N个数字的八进制数，它是一个完美的平方。
例子：
Input: N = 1
Output: 4
Explanation:
4 is the largest 1 digit Octal number which is also perfect square
Input: N = 2
Output: 61
Explanation:
49 is the largest number which is a 2-Digit Octal Number and also a perfect square.
Therefore 49 in Octal = 61
方法：
可以看出，在八进制数中也是最大的平方的最大数列是：
4, 61, 744, 7601, 77771, 776001 …..
众所周知，当数字大于8 k时，八进制系统中的数字会增加，其中k表示数字中的数字。因此，对于八进制数系统中的任何N位数字，其值都必须小于8 N + 1的值。因此，可以使用此观察结果得出的一般术语是–
N-Digit Octal Number = octal(pow(ceil(sqrt(pow(8, N))) -1, 2))
下面是上述方法的实现：
CPP
// C++ implementation to find the maximum
// N-digit octal number which is perfect square
#include
using namespace std;
// Function to convert decimal number
// to a octal number
void decToOctal(int n)
{
// Array to store octal number
int octalNum[100];
// Counter for octal number array
int i = 0;
while (n != 0) {
// Store remainder in
// octal array
octalNum[i] = n % 8;
n = n / 8;
i++;
}
// Print octal number array
// in reverse order
for (int j = i - 1; j >= 0; j--)
cout << octalNum[j];
cout << "\n";
}
void nDigitPerfectSquares(int n)
{
// Largest n-digit perfect square
int decimal = pow(
ceil(sqrt(pow(8, n))) - 1, 2
);
decToOctal(decimal);
}
// Driver Code
int main()
{
int n = 2;
nDigitPerfectSquares(n);
return 0;
} Java
// Java implementation to find the maximum
// N-digit octal number which is perfect square
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG
{
// Function to convert decimal number
// to a octal number
static void decToOctal(int n)
{
// Array to store octal number
int octalNum[] = new int[100];
// Counter for octal number array
int i = 0;
while (n != 0)
{
// Store remainder in
// octal array
octalNum[i] = n % 8;
n = n / 8;
i++;
}
// Print octal number array
// in reverse order
for (int j = i - 1; j >= 0; j--)
System.out.print(octalNum[j]);
System.out.println("\n");
}
static void nDigitPerfectSquares(int n)
{
// Largest n-digit perfect square
int decimal = (int) Math.pow(Math.ceil(Math.sqrt(Math.pow(8, n))) - 1, 2);
decToOctal(decimal);
}
// Driver code
public static void main (String[] args)
{
int n = 2;
nDigitPerfectSquares(n);
}
}
// This code is contributed by nidhivaPython3
# Python3 implementation to find the maximum
# N-digit octal number which is perfect square
from math import sqrt,ceil
# Function to convert decimal number
# to a octal number
def decToOctal(n) :
# Array to store octal number
octalNum = [0]*100;
# Counter for octal number array
i = 0;
while (n != 0) :
# Store remainder in
# octal array
octalNum[i] = n % 8;
n = n // 8;
i += 1;
# Print octal number array
# in reverse order
for j in range(i - 1, -1, -1) :
print(octalNum[j], end= "");
print();
def nDigitPerfectSquares(n) :
# Largest n-digit perfect square
decimal = pow(ceil(sqrt(pow(8, n))) - 1, 2);
decToOctal(decimal);
# Driver Code
if __name__ == "__main__" :
n = 2;
nDigitPerfectSquares(n);
# This code is contributed by AnkitRai01C#
// C# implementation to find the maximum
// N-digit octal number which is perfect square
using System;
class GFG
{
// Function to convert decimal number
// to a octal number
static void decToOctal(int n)
{
// Array to store octal number
int []octalNum = new int[100];
// Counter for octal number array
int i = 0;
while (n != 0)
{
// Store remainder in
// octal array
octalNum[i] = n % 8;
n = n / 8;
i++;
}
// Print octal number array
// in reverse order
for (int j = i - 1; j >= 0; j--)
Console.Write(octalNum[j]);
Console.WriteLine();
}
static void nDigitPerfectSquares(int n)
{
// Largest n-digit perfect square
int _decimal = (int) Math.Pow(Math.Ceiling(Math.Sqrt(Math.Pow(8, n))) - 1, 2);
decToOctal(_decimal);
}
// Driver code
public static void Main()
{
int n = 2;
nDigitPerfectSquares(n);
}
}
// This code is contributed by AnkitRai01Javascript
输出：
61