求数列的第 n 项，其中每个项 f[I]= f[I–1]–f[I–2]
原文:https://www . geesforgeks . org/find-the-n-term-of-series-where-每个 term-fi-fi-1-fi-2/
给定三个整数 X 、 Y 和 N ，任务是找到系列f[I]= f[I–1]–f[I–2]、 i > 1 的 N 第 项，其中 f[0] = X 和 f[1] = Y 。 例:
输入: X = 2，Y = 3，N = 3 输出: -2 该系列将为 2 3 1 -2 -3 -1 2 和 f[3] = -2 输入: X = 3，Y = 7，N = 8 输出: 4
方法:这里的一个重要观察是，在序列开始自我重复之前，会有 6 个不同的术语。因此，找到该系列的前 6 个术语，然后第 N 第T5】术语将与第 (N % 6) 第T9】术语相同。 以下是上述方法的实施:****
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the Nth term
// of the given series
int findNthTerm(int x, int y, int n)
{
int f[6];
// First and second term of the series
f[0] = x;
f[1] = y;
// Find first 6 terms
for (int i = 2; i <= 5; i++)
f[i] = f[i - 1] - f[i - 2];
// Return the Nth term
return f[n % 6];
}
// Driver code
int main()
{
int x = 2, y = 3, n = 3;
cout << findNthTerm(x, y, 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.io.*;
class GFG
{
// Function to find the nth term of series
static int findNthTerm(int x, int y, int n)
{
int[] f = new int[6];
f[0] = x;
f[1] = y;
// Loop to add numbers
for (int i = 2; i <= 5; i++)
f[i] = f[i - 1] - f[i - 2];
return f[n % 6];
}
// Driver code
public static void main(String args[])
{
int x = 2, y = 3, n = 3;
System.out.println(findNthTerm(x, y, n));
}
}
// This code is contributed by mohit kumar 29
Python 3
# Python3 implementation of the approach
# Function to return the Nth term
# of the given series
def findNthTerm(x, y, n):
f = [0] * 6
# First and second term of
# the series
f[0] = x
f[1] = y
# Find first 6 terms
for i in range(2, 6):
f[i] = f[i - 1] - f[i - 2]
# Return the Nth term
return f[n % 6]
# Driver code
if __name__ == "__main__":
x, y, n = 2, 3, 3
print(findNthTerm(x, y, n))
# This code is contributed by
# Rituraj Jain
C
// C# implementation of the approach
using System;
class GFG
{
// Function to find the nth term of series
static int findNthTerm(int x, int y, int n)
{
int[] f = new int[6];
f[0] = x;
f[1] = y;
// Loop to add numbers
for (int i = 2; i <= 5; i++)
f[i] = f[i - 1] - f[i - 2];
return f[n % 6];
}
// Driver code
public static void Main()
{
int x = 2, y = 3, n = 3;
Console.WriteLine(findNthTerm(x, y, n));
}
}
// This code is contributed by Ryuga
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
//PHP implementation of the approach
// Function to find the nth term of series
function findNthTerm($x, $y, $n)
{
$f = array(6);
$f[0] = $x;
$f[1] = $y;
// Loop to add numbers
for ($i = 2; $i <= 5; $i++)
$f[$i] = $f[$i - 1] - $f[$i - 2];
return $f[$n % 6];
}
// Driver code
$x = 2; $y = 3; $n = 3;
echo(findNthTerm($x, $y, $n));
// This code is contributed by Code_Mech.
?>
java 描述语言
<script>
// JavaScript implementation of the approach
// Function to return the Nth term
// of the given series
function findNthTerm(x, y, n)
{
let f = new Array(6);
// First and second term of the series
f[0] = x;
f[1] = y;
// Find first 6 terms
for (let i = 2; i <= 5; i++)
f[i] = f[i - 1] - f[i - 2];
// Return the Nth term
return f[n % 6];
}
// Driver code
let x = 2, y = 3, n = 3;
document.write(findNthTerm(x, y, n));
// This code is contributed by Surbhi Tyagi
</script>
Output:
-2
时间复杂度: O(1)
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