SALE - Editorial

editorial
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PROBLEM LINK:

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

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

DIFFICULTY:

Cakewalk

PREREQUISITES:

None

PROBLEM:

Chef goes to the supermarket to buy some items. Luckily there’s a sale going on under which Chef gets the following offer:

  • If Chef buys 3 items then he gets the item (out of those 3 items) having the lowest price as free.

For e.g. if Chef bought 3 items with the cost 6, 2 and 4, then he would get the item with cost 2 as free. So he would only have to pay the cost of the other two items which will be 6 + 4 = 10.

Chef buys 3 items having prices A, B and C respectively. What is the amount of money Chef needs to pay?

EXPLANATION:

We first need to find out the lowest price among A, B and C.
If A is lowest, then Chef needs to pay B + C. Similarly, if B is lowest, Chef will pay A+C and if C is lowest, Chef will pay A+B.

To find out if A is lowest, we can use two if conditions - (A \leq B) AND (A \leq C). We can have similar conditions for B and C.

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 , c ;
        cin >> a >> b >> c ;

        if(a <= b && a <= c)
        {
            cout << b+c << '\n' ;
        }
        else
        {
            if(b <= a && b <= c)
                cout << a+c << '\n' ;
            else
                cout << a+b << '\n' ;
        }
    }

    return 0;
}