# PROBLEM LINK:

* Author:* Abhishek Kumar

*Shekhar Srivastav*

**Tester:***Abhishek Kumar*

**Editorialist:****DIFFICULTY:** - CAKEWALK

**PREREQUISITES** : NONE

**PROBLEM SUMMARY:**

Difference of inputted number with result of

- swap(ith position , (l-i+1)th position) for all i=[1,l/2]

**HINT**

swap in a number of 1st and last , 2nd and 2nd last , so on is nothing but its reverse of number.

2 case arises

- if length of number is odd - n/2 swap pair will be there but (n/2)+1 will not , but still swap of (n/2)+1 with itself will not diturb our solution.
- if length of number is even- there will be exactly n/2 pairs .

so in generel approach we can reverse number without thinking about their length because its not going affect the solution.

**CPP SOLUTION**

```
using namespace std;
int rev(int n){
int sums=0;
while(n!=0) {
sums=sums*10+n%10;
n/=10;
}
return sums;
}
int main() {
int tc;
cin>>tc;
while(tc--){
int n;
cin>>n;
cout<<n-rev(n)<<"\n";
}
}
```