HDU-4925 Apple Tree

http://acm.hdu.edu.cn/showproblem.php?pid=4925

Apple Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 120    Accepted Submission(s):
82

Problem Description

I’ve bought an orchard and decide to plant some apple
trees on it. The orchard seems like an N * M two-dimensional map. In each grid,
I can either plant an apple tree to get one apple or fertilize the soil to speed
up its neighbors’ production. When a grid is fertilized, the grid itself doesn’t
produce apples but the number of apples of its four neighbor trees will double
(if it exists). For example, an apple tree locates on (x, y), and (x - 1, y),
(x, y - 1) are fertilized while (x + 1, y), (x, y + 1) are not, then I can get
four apples from (x, y). Now, I am wondering how many apples I can get at most
in the whole orchard?

 

Input

The input contains multiple test cases. The number of
test cases T (T<=100) occurs in the first line of input.
For each test
case, two integers N, M (1<=N, M<=100) are given in a line, which denote
the size of the map.

 

Output

For each test case, you should output the maximum
number of apples I can obtain.

 

Sample Input

2

2 2

3 3

 

Sample Output

8

32

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 using namespace std;
  int n;
  int main()
 {
    int n,m,map[][],i,j,t;
      scanf("%d",&t);
     while(t--)
     {
         scanf("%d%d",&n,&m);
         memset(map,,sizeof(map));
       for(i=;i<=n;i++)
       {
         for(j=;j<=m;j++)
         {
             map[i][j]=-;
             if((i%==)&&(j%==))
                  map[i][j]=;
             if((i%==)&&(j%==))
                 map[i][j]=;
          }
       }
 
          int a,b;
          for(i=;i<=n;i++)
          for(j=;j<=m;j++)
          {
              if(map[i][j]==)
                {
                   if(map[i-][j]==-&&i->=)
                        {
                            map[i][j]*=;
 
                        }
                   if(map[i][j-]==-&&j->=)
                        {
                            map[i][j]*=;
 
                        }
                   if(map[i+][j]==-&&i+<=n)
                   {
                       map[i][j]*=;
 
                   }
 
                    if(map[i][j+]==-&&j+<=m)
                        {
                            map[i][j]*=;
 
                        }
                }
          }
          int ans=;
          for(i=;i<=n;i++)
            for(j=;j<=m;j++)
            {
                if(map[i][j]!=-)
                ans+=map[i][j];
            }
 
         printf("%d\n",ans);
 
     }
     return ;
 }

 

Source

2014
Multi-University Training Contest 6