UVA11044 - Searching for Nessy

写的忘了一个问题，整除和余数。

题目：

Searching for Nessy

 

  Searching for Nessy 

The Loch Ness Monsteris a mysterious and unidentified animal said to inhabit Loch Ness, 
a large deep freshwater loch near the city of Inverness in northern Scotland. Nessie is usually categorized as a type of lake monster.
http://en.wikipedia.org/wiki/Loch_Ness_Monster

 

 


In July 2003, the BBC reported an extensive investigation of Loch Ness by a BBC team, using 600 separate sonar beams, found no trace of any ¨sea monster¨ (i.e., any large animal, known or unknown) in the loch. The BBC team concluded that Nessie does not exist. Now we want to repeat the experiment.

 

 


Given a grid of n rows and m columns representing the loch, 6$ \le$n, m$ \le$10000, find the minimum number s of sonar beams you must put in the square such that we can control every position in the grid, with the following conditions:

 

  • one sonar occupies one position in the grid; the sonar beam controls its own cell and the contiguous cells;
  • the border cells do not need to be controlled, because Nessy cannot hide there (she is too big).

For example,

 


$\textstyle \parbox{.5\textwidth}{ \begin{center} \mbox{} \epsfbox{p11044.eps} \end{center}}$$\textstyle \parbox{.49\textwidth}{ \begin{center} \mbox{} \epsfbox{p11044a.eps} \end{center}}$

 

\epsfbox{p11044b.eps}

where X represents a sonar, and the shaded cells are controlled by their sonar beams; the last figure gives us a solution.

 

Input 

The first line of the input contains an integer, t, indicating the number of test cases. For each test case, there is a line with two numbers separated by blanks, 6$ \le$n, m$ \le$10000, that is, the size of the grid (n rows and m columns).

 

Output 

For each test case, the output should consist of one line showing the minimum number of sonars that verifies the conditions above.

 

Sample Input 

 

 
3
6 6
7 7
9 13

 

Sample Output 

 

 
4
4
12

 

 

 

解答：

 

 1 #include<stdio.h>
 2 int main()
 3 {
 4     int t,n,m,x,y;
 5     scanf("%d",&t);
 6     while(t--)
 7     {
 8         scanf("%d%d",&n,&m);
 9         n-=2;
10         m-=2;
11         x=n/3+(n%3!=0);
12         y=m/3+(m%3!=0);
13         int i,k=0;
14         for(i=0;;i++)
15         {
16             if(m<=0&&n<=0)
17                 break;
18             else if(m<=0)
19             {
20                 n-=3;
21                 k+=y;
22             }
23             else if(n<=0)
24             {
25                 m-=3;
26                 k+=x;
27             }
28             else
29             {
30                 m-=3;
31                 n-=3;
32                 k+=(2*i+1);
33             }
34         }
35         printf("%d\n",k);
36     }
37     return 0;
38 }

 

 

 

posted @ 2013-03-02 21:50  上白泽慧音  阅读(218)  评论(0)  编辑  收藏  举报