**Problem Link:**

[CONTEST][1]

[PRACTICE][3]

**Author-** [Sameer Taneja][4]

**Tester-** [Vipin Khushu][2]

**Editorialist-** [Sameer Taneja][4]

**Difficulty:**

Simple

**Prerequisites:**

Easy

**Problem:**

We have to check nth element of unique series is positive or not? Where unique series just like fibbonaci series **u(n)=u(n-2)-u(n-1)**

**Explanation:**

Firstly we have to find nth unique number of the series .we can find this simple recursion or simple loop with 2-3 variables where a=100,b=97 initially.

For(i=2;i<n;i++){

nbsp;nbsp; c=a-b;

nbsp;nbsp;a=b; b=c;

}

Recurrsion:

Uni(int n){

If (n==1) return 100;

Else if (n==2)return 97;

Else return Uni(n-2)-Uni(n-1);

}

But it takes a lot time to calculate nth number in series as testcases< 10^5 and n< 10^5.it will lead to TLE in problem so we use an array to store unique series upto 10^5 elements. And can excess it at any time according to testcases.There will be no problem of TLE in it as we just have to apply a single loop upto 10^5 only.

For(i=2;i<n;i++){

U[n]=U[n-2]-U[n-1];} where U[0]=100 and U[1]=97

Now we check nth number is positive or not with a conditional statement.

**Solution:**

http://ideone.com/cC1vRN

[1]: https://www.codechef.com/CDWR2016/problems/FEST10

[2]: http://www.codechef.com/users/vipinkhushu

[4]: http://www.codechef.com/users/sameer87

[5]: http://www.codechef.com/users/vipinkhushu

[3]: https://www.codechef.com/problems/FEST10