FOR PEOPLE GETTING WA IN SECOND SUBTASK :

This is due to precision of log2 function.

try putting 2^56-1 as input in your code . u will see it gives 56 instead of 55. so u just need to raise the log2 value obtained to get the number again and compare it with your original number . since , we took floor of the log value , the number obtained by raising the log2 value must me smaller than original number ( or equal ) , if it is not just decrement your log2 value obtained by 1.

Refer this : https://www.codechef.com/viewsolution/26405550

Nothing to be ashamed of.

A lesson to be learned, if a solution involving built in function works for small numbers but fails on larger numbers, suspect the built in function!

I cant solve this problem even if I solved DOOFST

Can anyone help, i am getting WA for large inputs

#include<bits/stdc++.h>

#include

#define ll long long

using namespace std;

ll power(ll x, ll y, ll mod){

ll result = 1;

x = x % mod;

while(y > 0){

if(y&1){

result = (result *x)%mod;
}
x = (x* x)%mod;

y = y>>1;

}

return result;

}

int main(){

ll t; cin>>t;

int first = 0, second = 1, third;

vector series;

series.push_back(first);

series.push_back(second);

third = first+second;

first = second;

second = third;

series.push_back(third);

while(true){

```
first = second%10;
second = third%10;
if(first == 0 && second == 1){
break;
}
third = (first+second)%10;
series.push_back(third);
```

}

ll len = series.size()-2;

/*for(int i=0;i<series.size();i++){

cout<<series[i]<<" ";

}

cout<<endl;

cout<<"LEN "<<len<<endl; */

while(t–){

ll n; cin>>n;

ll a = floor(log2(n));

ll term = power(2,a,len);

//cout<<"TERM "<<term<<endl;

cout<<series[term-1]<<endl;

}

return 0;

}

this is link to my solution https://www.codechef.com/viewsolution/26495235 i used the logic that unit digit of fibonacci repeats after 60 digits and the selected element in array can be represented as a[2^n-1].

please tell me if code could be improved in some way

i also think the same, i think apart from the solution , its more challenging for a beginner to understand what mathematical concepts have been used , the editorial maker directly uses the various maths concept and make it harder , for beginners like me to understand .

so i think they should , go with more simpler approach to explain the logic.

another alternative is to use log2l instead of log2

It worked for me

Receiving RE for the third input. I have tried so many approaches but still issue persists.

```
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{
int t;
int f[] = {0,1,1,2,3,5,8,3,1,4,5,9,4,3,7,0,7,7,4,1,5,6,1,7,8,5,3,8,1,9,0,9,9,8,7,5,2,7,9,6,5,1,6,7,3,0,3,3,6,9,5,4,9,3,2,5,7,2,9,1};
Scanner in = new Scanner(System.in);
t = in.nextInt();
while(t>0)
{
long n,i;
i = 1;
n = in.nextInt();
while(i<=n)
{
i = i*2;
}
i = i/2;
i=(i-1)%60;
System.out.println(f[(int)i]);
t--;
}
}```
Is there any way we can find where im getting RE?
```

Please format your code - the forum software has mangled it and it is no longer valid Java!

Edit:

Try the testcase

```
1
576460752303423487
```

Thanks, prblem is with the way Im reading the input. changed “nextInt()” to “nextLong()” and it worked fine.

where am i going wrong?

arr =[0,1]

for x in range(60-2):

arr.append((arr[-1]+arr[-2])%10)

def red(arr,n):

brr = arr[:n]

while(len(brr)!=1):

brr = brr[1::2]

return brr[0]

tc = int(input())

for i in range(tc):

num = int(input())

print(red(arr,num%60))

plss tell me where its exceeding time limit

Can someone point out the error in my code?

```
import java.util.*;
class Fibnocci {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
int t,n,ans;
ArrayList<Integer>a = new ArrayList<Integer>();
ArrayList<Integer>e = new ArrayList<Integer>();
Scanner sc = new Scanner(System.in);
t = sc.nextInt();
while(t>0){
n = sc.nextInt();
a.add(0);
a.add(1);
for(int i=2;i<n;i++){
a.add(a.get(i-1)+a.get(i-2));
}
for(int i=0;i<n;i++){
a.set(i, a.get(i)%10);
}
if(a.get(a.size()-1)==1){
while(a.size()>1){
for(int i=0;i<a.size();i++){
if(i%2==0){
a.set(i,-1);
}
}
for(int i=0;i<a.size();i++){
if(a.get(i)==-1){
a.remove(i);
}
}
}
}
ans=a.get(0);
System.out.println(ans);
t--;
}
}
}
```

For people who need more clear explanation of the editorial can go to the GitHub link: Easy Fibonacci

can anyone check why i am getting wrong answer on this::

double n = in.readDouble();

int x = (int)(Math.log(n) /

Math.log(2));

x = (int)Math.pow(2, x);

out.printLine(fib(x-1)%10);

out.flush();

//fib(x-1) is implemented in logn using matrix expo