toj_2909.cpp

Tianjin University Online Judge 的80多道题目 .

/*2909.   Black and white painting Time Limit: 1.0 Seconds   Memory Limit: 65536K
Total Runs: 29   Accepted Runs: 23


You are visiting the Centre Pompidou which contains a lot of modern paintings. In particular you notice one painting which consists solely of black and white squares, arranged in rows and columns like in a chess board (no two adjacent squares have the same colour). By the way, the artist did not use the tool of problem A to create the painting.

Since you are bored, you wonder how many 8 × 8 chess boards are embedded within this painting. The bottom right corner of a chess board must always be white.
Input

The input contains several test cases. Each test case consists of one line with three integers n, m and c. (8 ≤ n, m ≤ 40000), where n is the number of rows of the painting, and m is the number of columns of the painting. c is always 0 or 1, where 0 indicates that the bottom right corner of the painting is black, and 1 indicates that this corner is white.

The last test case is followed by a line containing three zeros.
Output

For each test case, print the number of chess boards embedded within the given painting.
Sample Input

8 8 0
8 8 1
9 9 1
40000 39999 0
0 0 0

Sample Output

0
1
2
799700028



Source: University of Ulm Local Contest 2007*/
#include<cstdio>

int addLine( int num , int c )
{
  if ( num % 2 == 0 )
    return num / 2 - 4 + c;
  else
    return ( num - 7 ) / 2;
}  

int find( int n , int m , int c )
{
  int addition , additionM , additionN;
  if ( n <= 8 && m <= 8 )
    {
      if ( c == 0 )
	{
	  return 0;
	}
      else
	{
	  return 1;
	}
    }
  else if ( n <= 8 && m > 8 )
    {
      return ( n , m - 1 , 1 - c ) + addLine( m , c );
    }
  else if ( m <= 8 && n > 8 )
    {
      return ( n - 1 , m , 1 - c ) + addLine( n , c );
    }
  else if ( m > 8 && n > 8 )
    {
      addition = addLine( m , c ) + addLine( n , c ) - c;
      return( n - 1 , m - 1 , c ) + addition;
    }
}
int main()
{
  int n , m , c , result;
  while ( scanf( "%d%d%d"  ,&n , &m , &c ) != EOF && ( n != 0 || m != 0 || c != 0 ) )
    {
      result = find( n , m , c );
      printf( "%d\n" , result );
    }
  return 0;
}