**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 :

` ````
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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