打印下 Wythoff 序列的前 N 项
原文:https://www . geesforgeks . org/print-first-n-terms-lower-wythoff-sequence/
给定一个整数 N ，任务是打印下 Wythoff 序列的第一个 N 术语。
示例:
输入: N = 5 输出: 1，3，4，6，8
输入: N = 10 输出: 1、3、4、6、8、9、11、12、14、16
方法:下 Wythoff 序列是这样一个序列，其nthT5】项为 a(n) =地板(n φ)，其中φ=(1+sqrt(5))/2。所以，我们运行一个循环，找到序列的第一个 n 项。*
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 print the first n terms
// of the lower Wythoff sequence
void lowerWythoff(int n)
{
// Calculate value of phi
double phi = (1 + sqrt(5)) / 2.0;
// Find the numbers
for (int i = 1; i <= n; i++) {
// a(n) = floor(n * phi)
double ans = floor(i * phi);
// Print the nth numbers
cout << ans;
if (i != n)
cout << ", ";
}
}
// Driver code
int main()
{
int n = 5;
lowerWythoff(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 print the first n terms
// of the lower Wythoff sequence
static void lowerWythoff(int n)
{
// Calculate value of phi
double phi = (1 + Math.sqrt(5)) / 2.0;
// Find the numbers
for (int i = 1; i <= n; i++)
{
// a(n) = floor(n * phi)
double ans = Math.floor(i * phi);
// Print the nth numbers
System.out.print((int)ans);
if (i != n)
System.out.print(" , ");
}
}
// Driver code
public static void main(String[] args)
{
int n = 5;
lowerWythoff(n);
}
}
// This code is contributed by 29AjayKumar
Python 3
# Python3 implementation of the approach
# from math import sqrt,floor
from math import sqrt, floor
# Function to print the first n terms
# of the lower Wythoff sequence
def lowerWythoff(n) :
# Calculate value of phi
phi = (1 + sqrt(5)) / 2;
# Find the numbers
for i in range(1, n + 1) :
# a(n) = floor(n * phi)
ans = floor(i * phi);
# Print the nth numbers
print(ans,end="");
if (i != n) :
print( ", ",end = "");
# Driver code
if __name__ == "__main__" :
n = 5;
lowerWythoff(n);
# This code is contributed by AnkitRai01
C
// C# implementation of the approach
using System;
class GFG
{
// Function to print the first n terms
// of the lower Wythoff sequence
static void lowerWythoff(int n)
{
// Calculate value of phi
double phi = (1 + Math.Sqrt(5)) / 2.0;
// Find the numbers
for (int i = 1; i <= n; i++)
{
// a(n) = floor(n * phi)
double ans = Math.Floor(i * phi);
// Print the nth numbers
Console.Write((int)ans);
if (i != n)
Console.Write(" , ");
}
}
// Driver code
static public void Main ()
{
int n = 5;
lowerWythoff(n);
}
}
// This code is contributed by ajit.
java 描述语言
<script>
// Javascript implementation of above approach
// Function to print the first n terms
// of the lower Wythoff sequence
function lowerWythoff(n)
{
// Calculate value of phi
var phi = (1 + Math.sqrt(5)) / 2.0;
// Find the numbers
for(var i = 1; i <= n; i++)
{
// a(n) = floor(n * phi)
var ans = Math.floor(i * phi);
// Print the nth numbers
document.write( ans);
if (i != n)
document.write(", ");
}
}
// Driver code
var n = 5;
lowerWythoff(n);
// This code is contributed by SoumikMondal
</script>
Output:
1, 3, 4, 6, 8
版权属于：月萌API www.moonapi.com，转载请注明出处
