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.
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
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 using namespace std; 5 int n; 6 int main() 7 { 8 int n,m,map[200][200],i,j,t; 9 scanf("%d",&t); 10 while(t--) 11 { 12 scanf("%d%d",&n,&m); 13 memset(map,0,sizeof(map)); 14 for(i=1;i<=n;i++) 15 { 16 for(j=1;j<=m;j++) 17 { 18 map[i][j]=-1; 19 if((i%2==1)&&(j%2==1)) 20 map[i][j]=1; 21 if((i%2==0)&&(j%2==0)) 22 map[i][j]=1; 23 } 24 } 25 26 int a,b; 27 for(i=1;i<=n;i++) 28 for(j=1;j<=m;j++) 29 { 30 if(map[i][j]==1) 31 { 32 if(map[i-1][j]==-1&&i-1>=1) 33 { 34 map[i][j]*=2; 35 36 } 37 if(map[i][j-1]==-1&&j-1>=1) 38 { 39 map[i][j]*=2; 40 41 } 42 if(map[i+1][j]==-1&&i+1<=n) 43 { 44 map[i][j]*=2; 45 46 } 47 48 if(map[i][j+1]==-1&&j+1<=m) 49 { 50 map[i][j]*=2; 51 52 } 53 } 54 } 55 int ans=0; 56 for(i=1;i<=n;i++) 57 for(j=1;j<=m;j++) 58 { 59 if(map[i][j]!=-1) 60 ans+=map[i][j]; 61 } 62 63 printf("%d ",ans); 64 65 } 66 return 0; 67 }
Source