小于 N 的数字是完美立方体，它们的位数之和减少到一位数为 1
原文:https://www . geeksforgeeks . org/numbers-小于 n 的完美立方和它们的数字总和-简化为一位数-is-1/
给定一个数字 n ，任务是打印所有小于或等于 n 的数字，这些数字是完美立方体，它们的最终数字总和是 1 。 举例:
输入: n = 100 输出:1 64 64 = 6+4 = 10 = 1+0 = 1 输入: n = 1000 输出: 1 64 343 1000
方法:对于每一个小于或等于 n 的完美立方体，继续计算其位数之和，直到该数减少到一个位数(这里 O(1)方法)，如果该位数为 1，则打印完美立方体，否则跳到 n 下面的下一个完美立方体，直到所有完美立方体都被考虑。 以下是上述方法的实施:
C++
// C++ implementation of the approach
#include <cmath>
#include <iostream>
using namespace std;
// Function that returns true if the eventual
// digit sum of number nm is 1
bool isDigitSumOne(int nm)
{
//if reminder will 1
//then eventual sum is 1
if (nm % 9 == 1)
return true;
else
return false;
}
// Function to print the required numbers
// less than n
void printValidNums(int n)
{
int cbrt_n = (int)cbrt(n);
for (int i = 1; i <= cbrt_n; i++) {
int cube = pow(i, 3);
// If it is the required perfect cube
if (cube >= 1 && cube <= n && isDigitSumOne(cube))
cout << cube << " ";
}
}
// Driver code
int main()
{
int n = 1000;
printValidNums(n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG {
// Function that returns true if the eventual
// digit sum of number nm is 1
static boolean isDigitSumOne(int nm)
{
//if reminder will 1
//then eventual sum is 1
if (nm % 9 == 1)
return true;
else
return false;
}
// Function to print the required numbers
// less than n
static void printValidNums(int n)
{
int cbrt_n = (int)Math.cbrt(n);
for (int i = 1; i <= cbrt_n; i++) {
int cube = (int)Math.pow(i, 3);
// If it is the required perfect cube
if (cube >= 1 && cube <= n && isDigitSumOne(cube))
System.out.print(cube + " ");
}
}
// Driver code
public static void main(String args[])
{
int n = 1000;
printValidNums(n);
}
}
计算机编程语言
# Python3 implementation of the approach
import math
# Function that returns true if the eventual
# digit sum of number nm is 1
def isDigitSumOne(nm) :
#if reminder will 1
#then eventual sum is 1
if(nm % 9 == 1):
return True
else:
return False
# Function to print the required numbers
# less than n
def printValidNums(n):
cbrt_n = math.ceil(n**(1./3.))
for i in range(1, cbrt_n + 1):
cube = i * i * i
if (cube >= 1 and cube <= n and isDigitSumOne(cube)):
print(cube, end = " ")
# Driver code
n = 1000
printValidNums(n)
C
// C# implementation of the approach
using System;
class GFG
{
// Function that returns true if the
// eventual digit sum of number nm is 1
static bool isDigitSumOne(int nm)
{
//if reminder will 1
//then eventual sum is 1
if (nm % 9 == 1)
return true;
else
return false;
}
// Function to print the required
// numbers less than n
static void printValidNums(int n)
{
int cbrt_n = (int)Math.Ceiling(Math.Pow(n,
(double) 1 / 3));
for (int i = 1; i <= cbrt_n; i++)
{
int cube = (int)Math.Pow(i, 3);
// If it is the required perfect cube
if (cube >= 1 && cube <= n &&
isDigitSumOne(cube))
Console.Write(cube + " ");
}
}
// Driver code
static public void Main ()
{
int n = 1000;
printValidNums(n);
}
}
// This code is contributed by akt_mit
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
// Function that returns true if the
// eventual digit sum of number nm is 1
function isDigitSumOne($nm)
{
//if reminder will 1
//then eventual sum is 1
if ($nm % 9 == 1)
return true;
else
return false;
}
// Function to print the required numbers
// less than n
function printValidNums($n)
{
$cbrt_n = ceil(pow($n,1/3));
for ($i = 1; $i <= $cbrt_n; $i++)
{
$cube = pow($i, 3);
// If it is the required perfect cube
if ($cube >= 1 && $cube <= $n &&
isDigitSumOne($cube))
echo $cube, " ";
}
}
// Driver code
$n = 1000;
printValidNums($n);
// This code is contributed by Ryuga
?>
java 描述语言
<script>
// Javascript implementation of the approach
// Function that returns true if the
// eventual digit sum of number nm is 1
function isDigitSumOne(nm)
{
//if reminder will 1
//then eventual sum is 1
if (nm % 9 == 1)
return true;
else
return false;
}
// Function to print the required
// numbers less than n
function printValidNums(n)
{
let cbrt_n = Math.ceil(Math.pow(n, 1 / 3));
for (let i = 1; i <= cbrt_n; i++)
{
let cube = Math.pow(i, 3);
// If it is the required perfect cube
if (cube >= 1 && cube <= n && isDigitSumOne(cube))
document.write(cube + " ");
}
}
let n = 1000;
printValidNums(n);
</script>
Output:
1 64 343 1000
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