找出数列的第 n 项，其中每项交替相差 6 和 2
原文:https://www . geeksforgeeks . org/find-系列的第 n 个术语，其中每个术语交替地相差 6 和 2/
给定一个数字 N ，任务是找到数列的第N项，其中每个项交替相差 6 和 2。 例:
输入: N = 6 输出: 24 解释: 第 N 项为 0 + 6 + 2 + 6 + 2 + 6 + 2 = 24 输入: N = 3 输出: 14 解释: 第 N 项为 0 + 6 + 2 + 6 = 14
天真方法:想法是从 1 开始迭代，交替增加 6 和 2 ，直到我们到达第 n 个项。 以下是上述方法的实施:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find Nth term
void findNthTerm(int N)
{
int ans = 0;
// Iterate from 1 till Nth term
for (int i = 0; i < N; i++) {
// Check if i is even and
// then add 6
if (i % 2 == 0) {
ans = ans + 6;
}
// Else add 2
else {
ans = ans + 2;
}
}
// Print ans
cout << ans << endl;
}
// Driver Code
int main()
{
int N = 3;
findNthTerm(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 Nth term
static void findNthTerm(int N)
{
int ans = 0;
// Iterate from 1 till Nth term
for (int i = 0; i < N; i++) {
// Check if i is even and
// then add 6
if (i % 2 == 0) {
ans = ans + 6;
}
// Else add 2
else {
ans = ans + 2;
}
}
// Print ans
System.out.print(ans +"\n");
}
// Driver Code
public static void main(String[] args)
{
int N = 3;
findNthTerm(N);
}
}
// This code is contributed by PrinciRaj1992
Python 3
# Python3 program for the above approach
# Function to find Nth term
def findNthTerm(N):
ans = 0
# Iterate from 1 till Nth term
for i in range(N):
# Check if i is even and
# then add 6
if (i % 2 == 0) :
ans = ans + 6
# Else add 2
else :
ans = ans + 2
# Print ans
print(ans)
# Driver Code
if __name__=='__main__':
N = 3
findNthTerm(N)
# This code is contributed by AbhiThakur
C
// C# program for the above approach
using System;
class GFG{
// Function to find Nth term
static void findNthTerm(int N)
{
int ans = 0;
// Iterate from 1 till Nth term
for (int i = 0; i < N; i++) {
// Check if i is even and
// then add 6
if (i % 2 == 0) {
ans = ans + 6;
}
// Else add 2
else {
ans = ans + 2;
}
}
// Print ans
Console.Write(ans +"\n");
}
// Driver Code
public static void Main()
{
int N = 3;
findNthTerm(N);
}
}
// This code is contributed by AbhiThakur
java 描述语言
<script>
// JavaScript program for the above approach
// Function to find Nth term
function findNthTerm(N)
{
let ans = 0;
// Iterate from 1 till Nth term
for (let i = 0; i < N; i++) {
// Check if i is even and
// then add 6
if (i % 2 == 0) {
ans = ans + 6;
}
// Else add 2
else {
ans = ans + 2;
}
}
// Print ans
document.write(ans + "<br>");
}
// Driver Code
let N = 3;
findNthTerm(N);
// This code is contributed by Mayank Tyagi
</script>
Output:
14
时间复杂度:O(N) T5】高效方法:我们可以通过使用以下公式找到第 N 个**项:
- 如果 N 为奇数:第 N 项由(N/2 + 1)6 + (N/2)2 给出。
- 如果 N 为偶数:第 N 项由(N/2)6 + (N/2)2 给出。
以下是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find Nth term
void findNthTerm(int N)
{
int ans;
// Check if N is even
if (N % 2 == 0) {
// Formula for n is even
ans = (N / 2) * 6
+ (N / 2) * 2;
}
// Check if N is odd
else {
// Formula for N is odd
ans = (N / 2 + 1) * 6
+ (N / 2) * 2;
}
// Print ans
cout << ans << endl;
}
// Driver Code
int main()
{
int N = 3;
findNthTerm(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 Nth term
static void findNthTerm(int N)
{
int ans;
// Check if N is even
if (N % 2 == 0) {
// Formula for n is even
ans = (N / 2) * 6
+ (N / 2) * 2;
}
// Check if N is odd
else {
// Formula for N is odd
ans = (N / 2 + 1) * 6
+ (N / 2) * 2;
}
// Print ans
System.out.print(ans +"\n");
}
// Driver Code
public static void main(String[] args)
{
int N = 3;
findNthTerm(N);
}
}
// This code contributed by PrinciRaj1992
Python 3
# Python3 program for the above approach
# Function to find Nth term
def findNthTerm(N):
ans = 0;
# Check if N is even
if (N % 2 == 0):
# Formula for n is even
ans = (N // 2) * 6 + (N // 2) * 2;
# Check if N is odd
else:
# Formula for N is odd
ans = (N // 2 + 1) * 6 + (N // 2) * 2;
# Print ans
print(ans);
# Driver Code
if __name__ == '__main__':
N = 3;
findNthTerm(N);
# This code is contributed by Rajput-Ji
C
// C# program for the above approach
using System;
class GFG{
// Function to find Nth term
static void findNthTerm(int N)
{
int ans;
// Check if N is even
if (N % 2 == 0) {
// Formula for n is even
ans = (N / 2) * 6
+ (N / 2) * 2;
}
// Check if N is odd
else {
// Formula for N is odd
ans = (N / 2 + 1) * 6
+ (N / 2) * 2;
}
// Print ans
Console.Write(ans +"\n");
}
// Driver Code
public static void Main(String[] args)
{
int N = 3;
findNthTerm(N);
}
}
// This code is contributed by PrinciRaj1992
java 描述语言
<script>
// JavaScript program for the above approach
// Function to find Nth term
function findNthTerm( N)
{
let ans;
// Check if N is even
if (N % 2 == 0) {
// Formula for n is even
ans = parseInt(N / 2) * 6
+ parseInt(N / 2) * 2;
}
// Check if N is odd
else {
// Formula for N is odd
ans = parseInt(N / 2 + 1) * 6
+ parseInt(N / 2) * 2;
}
// Print ans
document.write(ans);
}
// Driver Function
// get the value of N
let N = 3;
// Calculate and print the Nth term
findNthTerm(N);
// This code is contributed by todaysgaurav
</script>
Output:
14
时间复杂度: O(1)
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