# Chess Match Codechef Solution

#### ByBrokenprogrammers

Dec 19, 2022

Hello Programmers In this post, you will know how to solve the Chess Match Codechef Solution.

One more thing to add, don’t directly look for the solutions, first try to solve the problems of Codechef by yourself. If you find any difficulty after trying several times, then you can look for solutions.

### Problem

In a Chess match “a + b”, each player has a clock which shows aa minutes at the start and whenever a player makes a move, bb seconds are added to this player’s clock. Time on a player’s clock decreases during that player’s turns and remains unchanged during the other player’s turns. If the time on some player’s clock hits zero (but not only in this case), this player loses the game.

There’s a 3 + 2 blitz chess match. After NN turns (i.e. ⌊N+12⌋⌊N+12⌋ moves made by white and ⌊N2⌋⌊N2⌋ moves made by black), the game ends and the clocks of the two players stop; they show that the players (white and black) have AA and BB seconds left respectively. Note that after the NN-th turn, b=2b=2 seconds are still added to the clock of the player that made the last move and then the game ends.

Find the total duration of the game, i.e. the number of seconds that have elapsed from the start of the game until the end.

### Input

• The first line of the input contains a single integer TT denoting the number of test cases. The description of TT test cases follows.
• The first and only line of each test case contains three space-separated integers NN, AA and BB.

### Output

For each test case, print a single line containing one integer — the duration of the game.

### Constraints

• 1≤T≤1051≤T≤105
• 10≤N≤10010≤N≤100
• 0≤A≤180+2⋅⌊N+12⌋0≤A≤180+2⋅⌊N+12⌋
• 0≤B≤180+2⋅⌊N2⌋0≤B≤180+2⋅⌊N2⌋
• for NN odd, A≥2A≥2
• for NN even, B≥2B≥2
• there is at least one valid game consistent with the input

```3
10 0 2
11 2 1
12 192 192
```

```378
379
```

### Explanation

Example case 1: The total time given to both clocks after 1010 turns is 2⋅(180+10)=3802⋅(180+10)=380 seconds. The total time left at the end is 0+2=20+2=2 seconds. The duration of the game is 380−2=378380−2=378 seconds.

Example case 3: The total time given to both clocks after 1212 turns is 2⋅(180+12)=3842⋅(180+12)=384 seconds. The total time left at the end is 192+192=384192+192=384 seconds. The duration of the game is 384−384=0384−384=0 seconds.

### Chess Match CodeChef Solution in JAVA

```import java.util.*;
import java.lang.*;
import java.io.*;
class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-->0){
int n = sc.nextInt();
int a = sc.nextInt();
int b = sc.nextInt();
int	T=2*(180+n);
int d=(T-(a+b));
System.out.println(d);
}
}
}```

### Chess Match CodeChef Solution in CPP

```#include <iostream>
using namespace std;
int main() {
int t,x,y;
cin>>t;
for(int i=0;i<t;i++)
{
int a,b,c;
cin>>a>>b>>c;
x=2*(180+a);
y=b+c;
cout<<(x-y)<<endl;
}
return 0;
}```

### Chess Match CodeChef Solution in Python

```test=int(input())
for j in range(test):
a,b,c=input().split()
a,b,c=int(a),int(b),int(c)
print(2*(180+a)-(b+c))```

Disclaimer: The above Problem (Chess Match) is generated by CodeChef but the solution is provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.

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