CareerCup Facebook Alex moving on a table

最新推荐文章于 2023-09-12 18:07:44 发布

转载最新推荐文章于 2023-09-12 18:07:44 发布 · 895 阅读

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本文详细探讨了在给定尺寸的表格中从左上角移动到右下角的路径问题，包括不同情况下的移动策略及计算公式。

Alex is standing on the top left cell (1,1) of a n*m table. The table has n rows and m columns. Initially, he is facing its right cell. He moves on the table in the following way:

>He moves one step forward.
>He turns to his right
>While moving forward, if he would go out of the table or reach a visited cell, he turns to his right.

He moves in the table as much as he can. Can you find out the number of cells he visits before he stops?

For example, given a 9x9 grid, the following would be his moves. The number on each cell represents the step he would land on that particular cell.
1 2 55 54 51 50 47 46 45
4 3 56 53 52 49 48 43 44
5 6 57 58 79 78 77 42 41
8 7 60 59 80 75 76 39 40
9 10 61 62 81 74 73 38 37
12 11 64 63 68 69 72 35 36
13 14 65 66 67 70 71 34 33
16 15 20 21 24 25 28 29 32
17 18 19 22 23 26 27 30 31

Input:
The first line of the input contains two integer numbers n and m.
n and m are between 1 and 100.

Output:
Print an integer to the output being the answer of the test.

Sample input #00:
3 3

Sample output #00:
9

Sample input #01:
7 4

Sample output #01:

--------------------------------------------------

if(n==1 || m==1) {
      cout<<(n+m-1) ; 
   }
   else if((n%2)&&(m%2)) {
    cout<<(n*m);
   }
   else if(n%2) {
       cout<<((n*m)-((n-2)*(m-2)));
    }
   else {
     cout<<(n*2);
   }

@Anonymous
Agree! But I guess you need to explain it a little bit...

Ok, the point is that, first assume n>=2 and m>=2, in this case you can convince yourself that the Alex has no difficulty moving to the bottom of the table, but
1) He can continue doing similar movement to the right if n is odd.
2) He is stuck at the bottom-left corner if n is even.
In case 2) the tiles he cover is 2*n; in case 1) he can move all the way to the right, then he is facing the situation just like before: 3) if m is odd he can continue to move upwards, 4) otherwise he is stuck in the bottom-right corner.
In case 3) the problem actually reduces to a smaller size problem if you rotate the board by 180 degrees, and Alex eventually can travel anywhere on the board.
In case 4) the number of tiles he travel is 2*n + 2*(m-2).

Finally, you need to take care the boundary cases when n==1 or m==1, which is not hard (see Anonymous' solution above).