找到修改后的斐波那契数列的第 n 个元素
原文:https://www . geeksforgeeks . org/find-修改后的斐波那契数列的第 n 个元素/
给定两个整数 A 和 B ，这是级数的前两项，另一个整数 N 。任务是使用斐波那契规则找到 N 第 个数字，即fib(I)= fib(I–1)+fib(I–2) 例:
输入: A = 2，B = 3，N = 4 输出: 8 级数为 2，3，5，8，13，21，… 第 4 个元素为 8。 输入: A = 5，B = 7，N = 10 输出: 343
方法:初始化变量 sum = 0 ，存储前两个值的总和。现在，运行从 i = 2 到 N 的循环，对于每个索引更新值 sum = A + B 和 A = B，B = sum 。最后，返回所需的第 n 个元素的总和。 以下是上述方法的实施:
C++
// C++ implementation of the approach
#include <iostream>
using namespace std;
// Function to return the Nth number of
// the modified Fibonacci series where
// A and B are the first two terms
int findNthNumber(int A, int B, int N)
{
// To store the current element which
// is the sum of previous two
// elements of the series
int sum = 0;
// This loop will terminate when
// the Nth element is found
for (int i = 2; i < N; i++) {
sum = A + B;
A = B;
B = sum;
}
// Return the Nth element
return sum;
}
// Driver code
int main()
{
int A = 5, B = 7, N = 10;
cout << findNthNumber(A, B, 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
import java.util.*;
class GFG
{
// Function to return the Nth number of
// the modified Fibonacci series where
// A and B are the first two terms
static int findNthNumber(int A, int B, int N)
{
// To store the current element which
// is the sum of previous two
// elements of the series
int sum = 0;
// This loop will terminate when
// the Nth element is found
for (int i = 2; i < N; i++)
{
sum = A + B;
A = B;
B = sum;
}
// Return the Nth element
return sum;
}
// Driver code
public static void main(String[] args)
{
int A = 5, B = 7, N = 10;
System.out.println(findNthNumber(A, B, N));
}
}
// This code is contributed by PrinciRaj1992
Python 3
# Python3 implementation of the approach
# Function to return the Nth number of
# the modified Fibonacci series where
# A and B are the first two terms
def findNthNumber(A, B, N):
# To store the current element which
# is the sum of previous two
# elements of the series
sum = 0
# This loop will terminate when
# the Nth element is found
for i in range(2, N):
sum = A + B
A = B
B = sum
# Return the Nth element
return sum
# Driver code
if __name__ == '__main__':
A = 5
B = 7
N = 10
print(findNthNumber(A, B, N))
# This code is contributed by Ashutosh450
C
// C# implementation of the approach
using System;
class GFG
{
// Function to return the Nth number of
// the modified Fibonacci series where
// A and B are the first two terms
static int findNthNumber(int A, int B, int N)
{
// To store the current element which
// is the sum of previous two
// elements of the series
int sum = 0;
// This loop will terminate when
// the Nth element is found
for (int i = 2; i < N; i++)
{
sum = A + B;
A = B;
B = sum;
}
// Return the Nth element
return sum;
}
// Driver code
public static void Main()
{
int A = 5, B = 7, N = 10;
Console.WriteLine(findNthNumber(A, B, N));
}
}
// This code is contributed by AnkitRai01
java 描述语言
<script>
// javascript implementation of the approach
// Function to return the Nth number of
// the modified Fibonacci series where
// A and B are the first two terms
function findNthNumber(A , B , N) {
// To store the current element which
// is the sum of previous two
// elements of the series
var sum = 0;
// This loop will terminate when
// the Nth element is found
for (i = 2; i < N; i++) {
sum = A + B;
A = B;
B = sum;
}
// Return the Nth element
return sum;
}
// Driver code
var A = 5, B = 7, N = 10;
document.write(findNthNumber(A, B, N));
// This code is contributed by todaysgaurav
</script>
Output:
343
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