# SPOJ - POWFIB (Fibo and non fibo) Time Limit Exceeds

Problem

Find (a^b)%M, where

a = Nth non-fibonacci number

b = Nth fibonacci number modulo M

M = 1000000007

Consider fibonacci series 1,1,2,3,.....

INPUT

First line contains T , the number of test cases.

Each next T lines contains a number N.

OUTPUT

Print T lines of output where each line corresponds to the required answer.

EXAMPLE

Input:

3

3

2

1

Output:

49

6

4

Constraints

1<=T<=100000

1<=N<=10^7

Here is my code for the problem link

#include <cstdio>
#include <unordered_map>
using namespace std;

typedef long long int ll;

ll M = 1000000007;

ll mulmod(ll a, ll b)       //modular multiplication
{
ll x = 0;
ll y = a%M;
while(b>0)
{
if(b%2)
x = (x+y)%M;
y = (y+y)%M;
b/=2;
}
return x%M;
}

ll modulo(ll a, ll b)       //modular exponentiation
{
ll x = 1;
ll y = a;
while(b)
{
if(b%2)
x = mulmod(x,y);
y = mulmod(y,y);
b/=2;
}
return x%M;
}

unordered_map<ll,ll> Fib;

ll fibo(ll n)       //n+1 th fibonacci number
{
if(n<2)
return 1;
if(Fib.find(n) != Fib.end())
return Fib[n];
Fib[n] = (fibo((n+1) / 2)*fibo(n/2) + fibo((n-1) / 2)*fibo((n-2) / 2)) % M;
return Fib[n];
}

ll nonfibo(ll n)        //nth non-fibonacci number
{
ll a = 1, b = 2, c = 3;
while(n>0)
{
a = b;
b = c;
c = a+b;
n-=(c-b-1);
}
n+=(c-b-1);
return n + b;
}

int main()
{
ll t;
scanf("%lld",&t);
while(t--)
{
ll n;
scanf("%lld",&n);
printf("%lld\n",modulo(nonfibo(n),fibo(n-1)));
}
return 0;
}


It exceeds the time limit, how do I improve the code?

• Repeatedly calculating fibonacci and non fibonacii cause time limit exceeded.Precompute the fibonacci and non fibonacci numbers in two arrays and use fast exponential multiplication method for power calculation.This will solve the problem. – aakansha Jul 22 '15 at 16:58

Here, you are using modular exponentiation and modular multiplication, which are fast enough. Your nonfibo function is also taking only O(lg(n)) time.
But your method for computing nth fibonacci number is not much efficient. Try using this Linear Recurrence Solving Method.