```
using namespace std;
int main() {
// your code goes here
int t,n;
int j =0;
cin>>t>>n;
while(t--){
int F[n];
if(n==1){F[0] = 0;}
if(n==2){F[1] = 1;
F[0] = 0;
}
for(int i = 0;i<n;i++){
if(i==0){
F[i] = 0;
}
else if(i ==1){
F[i] = 1;
}
else
{
F[i] = F[i-1] +F[i-2] ; }
}
int D[n],E[n/2+2];
for(int i = 0;i<n;i++){
D[i] = F[i]%10;
}
if(n == 1){
return 0;
}
else{
while(n >1){
j = 0;
for(int i =0;i<n;i++){
if(i%2){
E[j] = D[i];
j++;}
}n =j;
D[n];
for(int i =0;i<j;i++){
D[i] = E[i];
}
}
cout<<D[0];
}
}
return 0;
}
```

Please either format your code or link to your submission - the forum software has mangled it and it won’t compile!

Edit:

Assuming this is what you want to check - it gives no output for the testcase

```
1
1
```

1 Like

@ssjgz It is mentioned in challenge to end the process if l =1.

- Let D=(D1,D2,…,Dl)
- If l=1 the process ends.

Yes - the process ends, and then you have to *print* the number remaining