最大的 N 位八进制数，它是一个完美的正方形
给定一个自然数 N ，任务是找到最大的 N 位八进制数，它是一个完美的正方形。 例:
输入: N = 1 输出: 4 说明: 4 是最大的 1 位数八进制数，也是完美的平方 输入: N = 2 输出: 61 说明: 49 是最大的数，是 2 位数八进制数，也是完美的平方。 因此八进制 49 = 61
逼近: 可以观察到，八进制中也是正方的最大数级数是:
4, 61, 744, 7601, 77771, 776001 …..
正如我们所知，当一个数大于 8 k 时，八进制中的位数增加，其中 k 表示数中的位数。所以对于八进制数字系统中的任何 N 位数，都必须小于 8 N+1 的值。因此，利用这一观察可以得出的一般术语是–
N 位八进制数=八进制(幂(ceil(sqrt(幂(8，N))) -1，2))
以下是上述方法的实现:
卡片打印处理机（Card Print Processor 的缩写）
// C++ implementation to find the maximum
// N-digit octal number which is perfect square
#include <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)
// 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 nidhiva
Python 3
# 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 AnkitRai01
C
// 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 AnkitRai01
java 描述语言
<script>
// Javascript implementation to find the maximum
// N-digit octal number which is perfect square
// Function to convert decimal number
// to a octal number
function decToOctal(n)
{
// Array to store octal number
var octalNum = Array(100).fill(0);
// Counter for octal number array
var i = 0;
while (n != 0) {
// Store remainder in
// octal array
octalNum[i] = n % 8;
n = parseInt(n / 8);
i++;
}
// Print octal number array
// in reverse order
for (var j = i - 1; j >= 0; j--)
document.write(octalNum[j]);
document.write("<br>");
}
function nDigitPerfectSquares(n)
{
// Largest n-digit perfect square
var decimal = Math.pow(
Math.ceil(Math.sqrt(Math.pow(8, n))) - 1, 2
);
decToOctal(decimal);
}
// Driver Code
var n = 2;
nDigitPerfectSquares(n);
</script>
Output:
61
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