给定数字N ,任务是找到序列12、35、81、173、357,…中的第N个术语
例子:
Input: N = 2
Output: 35
2nd term = (12*2) + 11
= 35
Input: N = 5
Output: 357
5th term = (12*(2^4))+11*((2^4)-1)
= 357
方法:
- 每个数字都是通过将前一个数字乘以2并加上11来获得的。
- 由于起始编号为12。
1st term = 12
2nd term = (12 * 2) / 11 = 35
3rd term = (35 * 2) / 11 = 81
4th term = (81 * 2) / 11 = 173
And, so on....
- 通常,第N个数字可通过以下公式获得:
下面是上述方法的实现:
C++
// C++ program to find the Nth term
// in series 12, 35, 81, 173, 357, ...
#include
using namespace std;
// Function to find Nth term
int nthTerm(int N)
{
int nth = 0, first_term = 12;
// Nth term
nth = (first_term * (pow(2, N - 1)))
+ 11 * ((pow(2, N - 1)) - 1);
return nth;
}
// Driver Method
int main()
{
int N = 5;
cout << nthTerm(N) << endl;
return 0;
}
Java
// Java program to find the Nth term
// in series 12, 35, 81, 173, 357, ...
class GFG
{
// Function to find Nth term
static int nthTerm(int N)
{
int nth = 0, first_term = 12;
// Nth term
nth = (int) ((first_term * (Math.pow(2, N - 1)))
+ 11 * ((Math.pow(2, N - 1)) - 1));
return nth;
}
// Driver code
public static void main(String[] args)
{
int N = 5;
System.out.print(nthTerm(N) +"\n");
}
}
// This code is contributed by Rajput-Ji
Python3
# Python3 program to find the Nth term
# in series 12, 35, 81, 173, 357, ...
# Function to find Nth term
def nthTerm(N) :
nth = 0; first_term = 12;
# Nth term
nth = (first_term * (pow(2, N - 1))) + \
11 * ((pow(2, N - 1)) - 1);
return nth;
# Driver Method
if __name__ == "__main__" :
N = 5;
print(nthTerm(N)) ;
# This code is contributed by AnkitRai01
C#
// C# program to find the Nth term
// in series 12, 35, 81, 173, 357, ...
using System;
class GFG
{
// Function to find Nth term
static int nthTerm(int N)
{
int nth = 0, first_term = 12;
// Nth term
nth = (int) ((first_term * (Math.Pow(2, N - 1)))
+ 11 * ((Math.Pow(2, N - 1)) - 1));
return nth;
}
// Driver code
public static void Main(String[] args)
{
int N = 5;
Console.Write(nthTerm(N) +"\n");
}
}
// This code is contributed by PrinciRaj1992
Javascript
输出:
357