最接近 N 的斐波那契数
原文:https://www . geesforgeks . org/near-Fibonacci-number-to-n/
给定一个正整数 N ，任务是找到与给定整数 N 最近的斐波那契数。如果有两个斐波那契数与 N 有相同的差异，则打印较小的值。
示例:
输入: N = 20 输出: 21 说明:离 20 最近的斐波那契数是 21。
输入:N = 17 T3】输出: 13
方法:按照以下步骤解决问题:
- 如果 N 等于 0 ，则打印 0 作为结果。
- 初始化一个变量，比如说和，来存储离最近的斐波那契数。
- 初始化两个变量，将第一个设为 0 ，将第二个设为 1 ，存储斐波那契数列的第一项和第二项。
- 将第一个和第二个的总和存储在一个变量中，比如说第三个。
- 迭代至第三个的值最多为 N ，执行以下步骤:
- 将第一的值更新为第二和第二到第三的值。
- 将第一和第二的总和存储在变量第三中。
- 如果秒和 N 的绝对差最多是秒和 N 的值，那么更新 ans 的值为秒。
- 否则，将和的值更新为第三个。
- 完成上述步骤后，打印和的值作为结果。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the Fibonacci
// number which is nearest to N
void nearestFibonacci(int num)
{
// Base Case
if (num == 0) {
cout << 0;
return;
}
// Initialize the first & second
// terms of the Fibonacci series
int first = 0, second = 1;
// Store the third term
int third = first + second;
// Iterate until the third term
// is less than or equal to num
while (third <= num) {
// Update the first
first = second;
// Update the second
second = third;
// Update the third
third = first + second;
}
// Store the Fibonacci number
// having smaller difference with N
int ans = (abs(third - num)
>= abs(second - num))
? second
: third;
// Print the result
cout << ans;
}
// Driver Code
int main()
{
int N = 17;
nearestFibonacci(N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
class GFG{
// Function to find the Fibonacci
// number which is nearest to N
static void nearestFibonacci(int num)
{
// Base Case
if (num == 0)
{
System.out.print(0);
return;
}
// Initialize the first & second
// terms of the Fibonacci series
int first = 0, second = 1;
// Store the third term
int third = first + second;
// Iterate until the third term
// is less than or equal to num
while (third <= num)
{
// Update the first
first = second;
// Update the second
second = third;
// Update the third
third = first + second;
}
// Store the Fibonacci number
// having smaller difference with N
int ans = (Math.abs(third - num) >=
Math.abs(second - num)) ?
second : third;
// Print the result
System.out.print(ans);
}
// Driver Code
public static void main (String[] args)
{
int N = 17;
nearestFibonacci(N);
}
}
// This code is contributed by AnkThon
Python 3
# Python3 program for the above approach
# Function to find the Fibonacci
# number which is nearest to N
def nearestFibonacci(num):
# Base Case
if (num == 0):
print(0)
return
# Initialize the first & second
# terms of the Fibonacci series
first = 0
second = 1
# Store the third term
third = first + second
# Iterate until the third term
# is less than or equal to num
while (third <= num):
# Update the first
first = second
# Update the second
second = third
# Update the third
third = first + second
# Store the Fibonacci number
# having smaller difference with N
if (abs(third - num) >=
abs(second - num)):
ans = second
else:
ans = third
# Print the result
print(ans)
# Driver Code
if __name__ == '__main__':
N = 17
nearestFibonacci(N)
# This code is contributed by SURENDRA_GANGWAR
C
// C# program for the above approach
using System;
class GFG{
// Function to find the Fibonacci
// number which is nearest to N
static void nearestFibonacci(int num)
{
// Base Case
if (num == 0)
{
Console.Write(0);
return;
}
// Initialize the first & second
// terms of the Fibonacci series
int first = 0, second = 1;
// Store the third term
int third = first + second;
// Iterate until the third term
// is less than or equal to num
while (third <= num)
{
// Update the first
first = second;
// Update the second
second = third;
// Update the third
third = first + second;
}
// Store the Fibonacci number
// having smaller difference with N
int ans = (Math.Abs(third - num) >=
Math.Abs(second - num)) ?
second : third;
// Print the result
Console.Write(ans);
}
// Driver Code
public static void Main(string[] args)
{
int N = 17;
nearestFibonacci(N);
}
}
// This code is contributed by sanjoy_62
java 描述语言
<script>
// Javascript program for the above approach
// Function to find the Fibonacci
// number which is nearest to N
function nearestFibonacci(num)
{
// Base Case
if (num == 0) {
document.write(0);
return;
}
// Initialize the first & second
// terms of the Fibonacci series
let first = 0, second = 1;
// Store the third term
let third = first + second;
// Iterate until the third term
// is less than or equal to num
while (third <= num) {
// Update the first
first = second;
// Update the second
second = third;
// Update the third
third = first + second;
}
// Store the Fibonacci number
// having smaller difference with N
let ans = (Math.abs(third - num)
>= Math.abs(second - num))
? second
: third;
// Print the result
document.write(ans);
}
// Driver Code
let N = 17;
nearestFibonacci(N);
// This code is contributed by subhammahato348.
</script>
Output:
13
时间复杂度: O(log N) 辅助空间: O(1)
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