从 N 中减去最大完美立方体的次数
给定一个数字 N ，在每一步中，从 N 中减去最大的完美立方体(≤ N)。当 N > 0 时，重复这一步。任务是计算可以执行的步骤数。
示例:
输入: N = 100 输出: 4 第一步，100 –( 4 * 4 * 4)= 100–64 = 36 第二步，36 –( 3 * 3 * 3)= 36–27 = 9 第三步，9–(2 * 2 * 2)= 9–8 = 1 第四步，1–(1 * 1 * 1)= 1–1 = 0
输入: N = 150 输出: 5 第一步，150 –( 5 * 5 * 5)= 150–125 = 25 第二步，25 –( 2 * 2 * 2)= 25–8 = 17 第三步，17–(2 * 2 * 2)= 17–8 = 9 第四步，9–(2 * 2 * 2)= 9–8 = 1 第五步
接近:
- 得到一个数，从这个数中最大的完美立方体必须被缩小。
- 找到数字的立方根，并将结果转换为整数。数字的立方根可能在小数之后包含一些分数部分，这需要避免。
- 减去上一步中找到的整数的立方。这将从上面步骤中的数字中移除最大可能的完美立方体。
N = N - ((int) ∛N)3
- 用减少的数字重复上述两个步骤，直到它大于 0。
- 打印一个完美立方体从 n 减少的次数。这是最终结果。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the count of steps
int countSteps(int n)
{
// Variable to store the count of steps
int steps = 0;
// Iterate while N > 0
while (n) {
// Get the largest perfect cube
// and subtract it from N
int largest = cbrt(n);
n -= (largest * largest * largest);
// Increment steps
steps++;
}
// Return the required count
return steps;
}
// Driver code
int main()
{
int n = 150;
cout << countSteps(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 to return the count of steps
static int countSteps(int n)
{
// Variable to store the count of steps
int steps = 0;
// Iterate while N > 0
while (n > 0) {
// Get the largest perfect cube
// and subtract it from N
int largest = (int) Math.cbrt(n);
n -= (largest * largest * largest);
// Increment steps
steps++;
}
// Return the required count
return steps;
}
// Driver code
public static void main(String[] args)
{
int n = 150;
System.out.print(countSteps(n));
}
}
// This code is contributed by 29AjayKumar
Python 3
# Python3 implementation of the approach
from math import floor
# Function to return the count of steps
def countSteps(n):
# Variable to store the count of steps
steps = 0
# Iterate while N > 0
while (n):
# Get the largest perfect cube
# and subtract it from N
largest = floor(n**(1/3))
n -= (largest * largest * largest)
# Increment steps
steps += 1
# Return the required count
return steps
# Driver code
n = 150
print(countSteps(n))
# This code is contributed by mohit kumar 29
C
// C# implementation of the approach
using System;
class GFG{
// Function to return the count of steps
static int countSteps(int n)
{
// Variable to store the count of steps
int steps = 0;
// Iterate while N > 0
while (n > 0) {
// Get the largest perfect cube
// and subtract it from N
int largest = (int) Math.Pow(n,(double)1/3);
n -= (largest * largest * largest);
// Increment steps
steps++;
}
// Return the required count
return steps;
}
// Driver code
public static void Main(String[] args)
{
int n = 150;
Console.Write(countSteps(n));
}
}
// This code is contributed by PrinciRaj1992
java 描述语言
<script>
// JavaScript implementation of the approach
// Function to return the count of steps
function countSteps(n)
{
// Variable to store the count of steps
let steps = 0;
// Iterate while N > 0
while (n)
{
// Get the largest perfect cube
// and subtract it from N
let largest = Math.floor(Math.cbrt(n));
n -= (largest * largest * largest);
// Increment steps
steps++;
}
// Return the required count
return steps;
}
// Driver code
let n = 150;
document.write(countSteps(n));
// This code is contributed by Manoj.
</script>
Output:
5
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