Lucas序列中从1到N的数字总和

给定数字N。任务是找到卢卡斯序列中存在的从1到N的数字总和。

卢卡斯数按以下整数顺序表示：

2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123 ......

例子：

Input :  N = 10
Output : 17

Input : N = 5
Output : 10

方法：

• 遍历所有小于给定值N的卢卡斯数。
• 用0初始化sum变量。
• 继续添加这些卢卡斯数字以获得所需的总和。

下面是上述方法的实现：

C++

// C++ program to find sum of numbers from
// 1 to N which are in Lucas Sequence
#include 
using namespace std;
  
// Function to return the
// required sum
int LucasSum(int N)
{
    // Generate lucas number and keep on
    // adding them
    int sum = 0;
    int a = 2, b = 1, c;
  
    sum += a;
  
    while (b <= N) {
  
        sum += b;
        int c = a + b;
        a = b;
        b = c;
    }
  
    return sum;
}
  
// Driver code
int main()
{
    int N = 20;
    cout << LucasSum(N);
    return 0;
}

Java

// java program to find sum of numbers from
// 1 to N which are in Lucas Sequence
class GFG
{
  
// Function to return the
// required sum
static int LucasSum(int N)
{
    // Generate lucas number and keep on
    // adding them
    int sum = 0;
    int a = 2, b = 1, c;
  
    sum += a;
  
    while (b <= N) {
  
        sum += b;
        c = a + b;
        a = b;
        b = c;
    }
  
    return sum;
}
  
// Driver code
public static void main(String[] args)
{
    int N = 20;
    System.out.println(LucasSum(N));
      
}
// This code is contributed by princiraj1992 
}

Python3

# Python3 program to find Sum of 
# numbers from 1 to N which are 
# in Lucas Sequence
  
# Function to return the
# required Sum
def LucasSum(N):
      
    # Generate lucas number and 
    # keep on adding them
    Sum = 0
    a = 2
    b = 1
    c = 0
  
    Sum += a
  
    while (b <= N):
  
        Sum += b
        c = a + b
        a = b
        b = c
  
    return Sum
  
# Driver code
N = 20
print(LucasSum(N))
  
# This code is contributed 
# by mohit kumar

C#

// C# program to find sum of numbers from
// 1 to N which are in Lucas Sequence
using System;
  
class GFG
{
  
// Function to return the
// required sum
static int LucasSum(int N)
{
    // Generate lucas number and keep on
    // adding them
    int sum = 0;
    int a = 2, b = 1, c;
  
    sum += a;
  
    while (b <= N) 
    {
  
        sum += b;
        c = a + b;
        a = b;
        b = c;
    }
  
    return sum;
}
  
// Driver code
public static void Main(String[] args)
{
    int N = 20;
    Console.WriteLine(LucasSum(N));
}
}
  
// This code contributed by Rajput-Ji

输出：

46