Fibonacci Modified


Problem Statement :


Implement a modified Fibonacci sequence using the following definition:

Given terms t[i] and t[i+1] where i ∈ (1,∞), term t[i+2] is computed as:
t(i+2) = ti + t(i+2)^2

Given three integers, t1, t2, and n, compute and print the nth term of a modified Fibonacci sequence.

Example
t1 = 0
t2 =0
n =6

  t3 = 0+1^2 = 1
  t4 = 1+1^2 = 2
  t5 = 1+2^2 = 5
  t6 = 2+5^2 = 27
Return 27.

Function Description

Complete the fibonacciModified function in the editor below. It must return the nth number in the sequence.

fibonacciModified has the following parameter(s):

int t1: an integer
int t2: an integer
int n: the iteration to report

Returns

int: the nth number in the sequence
Note: The value of t[n] may far exceed the range of a 64-bit integer. Many submission languages have libraries that can handle such large results but, for those that don't (e.g., C++), you will need to compensate for the size of the result.

Input Format

A single line of three space-separated integers, the values of t1, t2, and n.

Constraints

0 <= t1, t2 <= 2
3 <= n <= 20
tn may far exceed the range of a 64-bit integer.


Solution :



title-img 


                            Solution in C :

In C++ :






#include <iostream>
#include <utility>

#include <boost/multiprecision/cpp_int.hpp>

using boost::multiprecision::cpp_int;

cpp_int fib(cpp_int a, cpp_int b, unsigned int n)
{
    for(unsigned int i = 2; i < n; ++i)
    {
        cpp_int temp = a + b*b;
        a = b;
        b = temp;
    }
    return b;
}

int main()
{
    unsigned int a, b, n;

    std::cin >> a >> b >> n;
    std::cout << fib(a, b, n);
}








In Java :






import java.math.BigInteger;
import java.util.Scanner;


public class Solution {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		BigInteger a = BigInteger.valueOf(in.nextInt());
		BigInteger b = BigInteger.valueOf(in.nextInt());
		int n = in.nextInt();
		for(int i = 2; i < n; i++) {
			BigInteger next = b.multiply(b).add(a);
			a = b;
			b = next;
		}
		System.out.println(b);
	}
}








In C :






#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

#define MAXL    26624

unsigned int MADD(unsigned int* pC, unsigned int* pB, unsigned int* pA, unsigned int n)
{
    unsigned int i,j,x;
    
    for (i=0; i<n; i++) pC[i] = pA[i];
    for (i=0; i<n; i++)
    {
        for (j=0; j<n; j++)
        {
            if ((x = (pC[i+j] += pB[i]*pB[j])) < 10000) continue;
            x /= 10000; pC[i+j+1] += x; pC[i+j] -= x*10000;            
        }
        if ((x = pC[i+j]) < 10000) continue;
        x /= 10000; pC[i+j+1] += x; pC[i+j] -= x*10000;            
    }
    n <<= 1; while (pC[n-1] == 0) n--;
    return n;
}

int main() {

    /* Enter your code here. Read input from STDIN. Print output to STDOUT */
    unsigned int N,m,n=1;
    unsigned int* p;
    unsigned int* pA;
    unsigned int* pB;
    unsigned int* pC;
    unsigned int A[MAXL];
    unsigned int B[MAXL];
    unsigned int C[MAXL];

    memset(pA=A, 0, sizeof(A));
    memset(pB=B, 0, sizeof(B));
    memset(pC=C, 0, sizeof(C));

    scanf("%d %d %d\n", pA, pB, &N);

    while (N-- > 2)
    {
        n = MADD(pC, pB, pA, n);
        p = pC; pC=pA; pA = pB; pB=p;
    }
    printf("%d", p[--n]);
    while (n > 0) printf("%04u", p[--n]);
    printf("\n");
    return 0;
}








In Python3 :






a, b, n = [int(x) for x in input().split(" ")]

if n == 0:
    print(a)
if n == 1:
    print(b)

for _ in range(n-2):
    a, b = b, b*b + a
    
print(b)








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