# GAME OF NUMBERS

Bob likes to play games very much. Today, Bob is playing a game in which the number A will be given and Bob has to guess if A will be even or odd before seeing the value of A.

• If he correctly guesses it then he wins B coins.
• Else He will lose B coins.

Initially, he has C coins.
Since Bob thinks of even numbers as lucky numbers, He will always guess A as an even number.
You have to tell the number of coins he will have after playing this game exactly one time. Input

## Format:

The first line of input will contain the number of test cases T.

• In the next T lines, Each line will be containing 3 integers A, B, C.
• Where A is the given value,
• B is the winning/losing amount
• C is the amount Bob is initially having.

## Output Format:

Print the value of the number of coins Bob will have after playing the game exactly one time.

## Constraints

• 1 ≤ T ≤ 100
• 1 ≤ A, B, C ≤ 100

2
1 3 5
4 4 5

2
9

## Explanation

In the first test case, the value of A is odd so Bob loses 3 coins. Initially, he was having 5 coins, hence he has 2 coins left
In the second test case, A is even so Bob wins 4 coins, so he has now a total of 9 coins.

# Solution

The approach here is to add in case of even input and subtract when it’s odd number.

``````#include <iostream>
using namespace std;
void techspiritss()
{
int a, b, c;
cin >> a >> b >> c;
if (a % 2) cout << c - b;
else cout << c + b;
cout << endl;
}
int main()
{
int t = 1;
cin >> t;
while (t--) techspiritss();
return 0;
}

``````