OJ Board

I don't know what exactly the solution is , but this is what I thought:

n,m are integers

Rook:the smaller of n,m
Knight: n*m/2
Queen: the smaller of n,m
King: (n/2)*(m*2)

I don't know why this code can't get accepted. It always gets WA. But I think the algorithm isn't wrong.Could someone find out the bug?
[c]
#include<stdio.h>
void main(void)
{
int i,x,m,n;
char chess;
scanf("%d",&i);
for(x=0;x<i;x++)
{
fflush(stdin);
scanf("%c %d %d",&chess,&m,&n);
if(chess=='r')
{
if(m>=n)
printf("%d\n",n);
else
printf("%d\n",m);
}
if(chess=='k')
{
if(m*n%2==1)
printf("%d\n",(m*n+1)/2);
else
printf("%d\n",m*n/2);
}
if(chess=='Q')
{
if(m>=n)
printf("%d\n",n);
else
printf("%d\n",m);
}
if(chess=='K')
{
if(m%2==1)
m=(m+1)/2;
else
m/=2;
if(n%2==1)
n=(n+1)/2;
else
n/=2;
printf("%d\n",m*n);
}
}
}
[/c]

put a space in scanf's formating before reading %c, or else it will read white spaces too.

btw:
if (a%2==1) b=(a+1)/2; else b=a/2
has the same result as only:
b=(a+1)/2

I finally got accepted, thanks.

Can anyone give me some critical input. I am trying this quite a some time. I am getting WA.

Thanks in advanced

For rook and queen it is the minimum of ( row, column).

For knight - picture the board as a grid consisting of black and white colours and count the number of black or white squares.

For king - do the same thing as knight but skip the alternate row.

Hope it helps.

I used this algorithm. But still WA. I think the Knigths causing me the problem. Can anyone help me with some input.

Thanks for the reply

HI Shantanu,

Here is some input that might come to a help;

k 5 5
k 4 4
k 5 4
k 4 5

output:

13
8
10
10

Hope it helps.

sohel wrote: For rook and queen it is the minimum of ( row, column).

I don't understand..say the board is 2 by 2. Then how can you possibly put 2 queens on the board so that they don't attack each other? Should the answer here not be 1?

[Edit] I am foolish..I didn't realize the problem constraints say that m and n are both greater than or equal to 4[/Edit]

#include<stdio.h>
#include<conio.h>

int find_min(int p,int q)
{
if(p>q)
return q;
else
return p;


}

int main(void)
{
long cs;
int m, n;
char chpcs;

scanf("%ld", &cs);

while(cs)
{
chpcs=getch();
scanf("%d %d",&m,&n);

if(chpcs=='r'||chpcs=='Q')
printf("%d\n", find_min(m,n));


else if(chpcs=='k')
printf("%d\n",m*n/2);

else
printf("%d\n",( ((m+1)/2) * ((n+1)/2) ) );


cs--;


}
return 0;
}

just solve this problem without using conio.h and getch().

I think you will be able to get AC after this

And remove your code after AC

the minimum possible size of the board is 4*4...

I used differents formulas for the King and the Knight.

the following code is giving wrong answer what is the problem?
#include<iostream>
using namespace std;
int main(void)
{
int testno,testcase=1,product,m,n;
char a;
cin>>testno;
while(testcase<=testno)
{

cin>>a>>m>>n;


product=1;
if(a=='r'||a=='q')
{
if(m<=n)cout<<m<<'\n';
else cout<<n<<'\n';
}
else if(a=='k')
{
if((m*n)%2==0)cout<<(m*n)/2<<'\n';
else cout<<(m*n)/2+1<<'\n';
}
else if(a=='K')
{
if((m%2)==0 )product*=(m/2);
else product*=(m/2)+1;
if((n%2)==0)product*=(n/2);
else product*=(n/2)+1;
cout<<product<<'\n';

}
testcase++;
}
return 0;
}

This will certainly help you:
output: happy programming!!