以前的完美正方形和小于 N 的立方体数量
原文:https://www . geeksforgeeks . org/previous-perfect-square-and-cube-number-小于 number-n/
给定一个整数 N ，任务是找到比数字 N 小的前一个完美正方形或完美立方体。 示例:
输入: N = 6 输出: 完美正方= 4 完美正方= 1 输入: N = 30 输出: 完美正方= 25 完美正方= 27
方法:小于 N 的前一个完美平方数可以计算如下:
- 求给定数 N 的平方根。
- 使用各自语言的楼层功能计算其楼层值。
- 然后减去 1，如果 N 已经是一个完美的正方形。
- 打印该数字的平方。
小于 N 的前一个完美立方数可以计算如下:
- 求给定 n 的立方根。
- 使用各自语言的楼层功能计算其楼层值。
- 然后减去 1，如果 N 已经是一个完美的立方体。
- 打印该数字的立方体。
下面是上述方法的实现:
C++
// C++ implementation to find the
// previous perfect square and cube
// smaller than the given number
#include <cmath>
#include <iostream>
using namespace std;
// Function to find the previous
// perfect square of the number N
int previousPerfectSquare(int N)
{
int prevN = floor(sqrt(N));
// If N is already a perfect square
// decrease prevN by 1.
if (prevN * prevN == N)
prevN -= 1;
return prevN * prevN;
}
// Function to find the
// previous perfect cube
int previousPerfectCube(int N)
{
int prevN = floor(cbrt(N));
// If N is already a perfect cube
// decrease prevN by 1.
if (prevN * prevN * prevN == N)
prevN -= 1;
return prevN * prevN * prevN;
}
// Driver Code
int main()
{
int n = 30;
cout << previousPerfectSquare(n) << "\n";
cout << previousPerfectCube(n) << "\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
// previous perfect square and cube
// smaller than the given number
import java.util.*;
class GFG{
// Function to find the previous
// perfect square of the number N
static int previousPerfectSquare(int N)
{
int prevN = (int)Math.floor(Math.sqrt(N));
// If N is already a perfect square
// decrease prevN by 1.
if (prevN * prevN == N)
prevN -= 1;
return prevN * prevN;
}
// Function to find the
// previous perfect cube
static int previousPerfectCube(int N)
{
int prevN = (int)Math.floor(Math.cbrt(N));
// If N is already a perfect cube
// decrease prevN by 1.
if (prevN * prevN * prevN == N)
prevN -= 1;
return prevN * prevN * prevN;
}
// Driver Code
public static void main(String[] args)
{
int n = 30;
System.out.println(previousPerfectSquare(n));
System.out.println(previousPerfectCube(n));
}
}
// This code is contributed by Rohit_ranjan
Python 3
# Python3 implementation to find the
# previous perfect square and cube
# smaller than the given number
import math
import numpy as np
# Function to find the previous
# perfect square of the number N
def previousPerfectSquare(N):
prevN = math.floor(math.sqrt(N));
# If N is already a perfect square
# decrease prevN by 1.
if (prevN * prevN == N):
prevN -= 1;
return prevN * prevN;
# Function to find the
# previous perfect cube
def previousPerfectCube(N):
prevN = math.floor(np.cbrt(N));
# If N is already a perfect cube
# decrease prevN by 1.
if (prevN * prevN * prevN == N):
prevN -= 1;
return prevN * prevN * prevN;
# Driver Code
n = 30;
print(previousPerfectSquare(n));
print(previousPerfectCube(n));
# This code is contributed by Code_Mech
C
// C# implementation to find the
// previous perfect square and cube
// smaller than the given number
using System;
class GFG{
// Function to find the previous
// perfect square of the number N
static int previousPerfectSquare(int N)
{
int prevN = (int)Math.Floor(Math.Sqrt(N));
// If N is already a perfect square
// decrease prevN by 1.
if (prevN * prevN == N)
prevN -= 1;
return prevN * prevN;
}
// Function to find the
// previous perfect cube
static int previousPerfectCube(int N)
{
int prevN = (int)Math.Floor(Math.Cbrt(N));
// If N is already a perfect cube
// decrease prevN by 1.
if (prevN * prevN * prevN == N)
prevN -= 1;
return prevN * prevN * prevN;
}
// Driver Code
public static void Main(String[] args)
{
int n = 30;
Console.WriteLine(previousPerfectSquare(n));
Console.WriteLine(previousPerfectCube(n));
}
}
// This code is contributed by sapnasingh4991
java 描述语言
<script>
// JavaScript implementation to find the
// previous perfect square and cube
// smaller than the given number
// Function to find the previous
// perfect square of the number N
function previousPerfectSquare(N)
{
let prevN = Math.floor(Math.sqrt(N));
// If N is already a perfect square
// decrease prevN by 1.
if (prevN * prevN == N)
prevN -= 1;
return prevN * prevN;
}
// Function to find the
// previous perfect cube
function previousPerfectCube(N)
{
let prevN = Math.floor(Math.cbrt(N));
// If N is already a perfect cube
// decrease prevN by 1.
if (prevN * prevN * prevN == N)
prevN -= 1;
return prevN * prevN * prevN;
}
// Driver Code
let n = 30;
document.write(previousPerfectSquare(n) + "<br>");
document.write(previousPerfectCube(n) + "<br>");
// This code is contributed by Manoj.
</script>
Output:
25
27
时间复杂度: O(sqrt(n))
辅助空间: O(1)
版权属于：月萌API www.moonapi.com，转载请注明出处
