Covid and Theatre Tickets Solution

Problem

Mr. Chef is the manager of the Code cinemas and after a long break, the theatres are now open to the public again. To compensate for the loss in revenue due to Covid-19, Mr. Chef wants to maximize the profits for every show from now on and at the same time follow the guidelines set the by government. The guidelines are:

• If two people are seated in the same row, there must be at least one empty seat between them.
• If two people are seated in different rows, there must be at least one completely empty row between them. That is, if there are people seated in rows $i$ and $j$ where $i \lt j$, there must be some row $k$ such that $i \lt k \lt j$ and nobody is seated in row $k$.

Given the information about the number of rows and the number of seats in each row, find the maximum number of tickets Mr. Chef can sell.

Input Format

• The first line of input will contain a single integer $T$, denoting the number of test cases. The description of $T$ test cases follows.
• Each test case consists of a single line of input containing two space-separated integers $N, M$ — the number of rows and the number of seats in each row, respectively.

Output Format

For each test case, output a single line containing one integer – the maximum number of tickets Mr. Chef can sell.

Constraints

• $1 \leq T \leq 100$
• $1 \leq N,M\leq 100$

Sample 1:

Input

Output

3
1 5
3 3
4 4

3
4
4

Explanation:

Test Case 1: There is only one row with five seats. Mr. Chef can sell a maximum of 3 tickets for seat numbers 1, 3 and 5.

Test Case 2: There are three rows with three seats each. Mr. Chef can sell a maximum of 4 tickets, for seats at the start and end of row numbers 1 and 3.

Test Case 3: There are four rows with four seats each. Mr. Chef can sell a maximum of 4 tickets, for example by choosing the seats at the start and end of row numbers 1 and 4.

Program :

 #include<stdio.h>

#include<math.h>

int main()

{

    int a,b=0,j,k,n,t,i,count;

    scanf("%d",&t);

    while(t--)

    {

        b=1,count=0;

        scanf("%d %d",&n,&k);

        for(j=1;j<=n;j=j+2)

        {

            b=1;

            while(b<=k)

            {

                b=b+2;

                count++;

            }

        }

    printf("%d\n",count);

    }

}