BLACKJACK - Editorial

editorial
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PROBLEM LINK:

Contest Division 1
Contest Division 2
Contest Division 3
Contest Division 4

Setter: Lavish Gupta
Tester: Abhinav Sharma, Aryan
Editorialist: Lavish Gupta

DIFFICULTY:

Cakewalk

PREREQUISITES:

None

PROBLEM:

Chef is playing a variant of Blackjack, where 3 numbers are drawn and each number lies between 1 and 10 (with both 1 and 10 inclusive). Chef wins the game when the sum of these 3 numbers is exactly 21.

Given the first two numbers A and B, that have been drawn by Chef, what should be 3-rd number that should be drawn by the Chef in order to win the game?

Note that it is possible that Chef cannot win the game, no matter what is the 3-rd number. In such cases, report -1 as the answer.

EXPLANATION:

Let us represent the 3-rd number that should be drawn by Chef by C. The following condition needs to be satisfied: A+B+C = 21.

This can be rewritten as C = 21 - (A+B). However, note that the 3-rd number can only lie between 1 and 10 (inclusive).

Hence, after getting the above value of C, we need to check if the resulting C is between 1 and 10. If Yes, we have got our answer as C, otherwise the answer is -1.

TIME COMPLEXITY:

O(1) for each test case.

SOLUTION:

Editorialist's Solution 
#include<bits/stdc++.h>
#define ll long long
using namespace std ;

int main()
{
    ll t;
    cin >> t ;
    while(t--)
    {
         ll a , b ;
        cin >> a >> b ;
        ll ans = (21 - (a+b)) ;

        if(ans >= 1 && ans <= 10)
            cout << ans << '\n' ;
        else
            cout << -1 << '\n' ;
    }

    return 0;
}