OJ Board

Input: 5 5 ..AAA E.BBB ..AA. CC.DD CC.D. 25 33 ................................. ................................. ................................. ................................. ................................. ................................. ....AAAAA.....AAAA.........AAAAA. ......AAAAAAAAA...

yayas wrote:

Code: Select all

printf("Tower : #%d\n",tm);

That is OK with the Output part of the problem, but their test output simply writes: I have got AC after removing colon and space.

Straight forward technique has already been shown: #define ROWS 10 #define COLS 5 int **ppnMatrix = (int**)calloc(ROWS, sizeof(int)); for (int i = 0; i < ROWS; i++) ppnMatrix[i] = (int*)calloc(COLS, sizeof(int)); // now destroy it: for (i = 0; i < ROWS; i++) free((void*)ppnMatrix[i]); free((void*)pp...

Sorry if I sounded rude, Adrian, I meant it's fishy that 95% of people didn't find out that both LIS and simple greedy are fundamentally incorrect. :wink: Well, it seems like I have got AC again, by modifying the greedy algorithm. It now uses construction approach: 1. start with an empty stack 2. in...

I've first got AC on greedy algorithm, and now it's not AC anymore (I haven't received email stating the new judgement yet). However, acceptance rate is now less than 5%... Something's fishy here.

This is not in fact a LIS problem, but rather one that is solved with greedy algorithm. Before using LIS, I suppose that you are sorting the input, boiling down to a greedy algorithm where sorting criterion is the cost function. I've got AC on greedy algo by maximizing "remaining strength"...

Well, I finally got it AC in only 0.002 sec, without use of big number division. To solve the problem, you got to transform the expression to the form: D(t, a, b) = sum(i=0, a/b - 1) (t^b)^i, where a%b=0 This sum can be further solved recursively: D(t, a, b) = f(t^b, a/b) f(x, 1) = 1 f(x, 2k) = (1 +...

I've got TLE as well, and here's what I did: 1. checked for special cases (b < a, b = 0, a = 0 -- although problem text says POSITIVE integers --, t = 1) - in each of these cases, result is either integer 0, or is not an integer 2. checked whether (a - b) * log10(t) > 99 - in this case result is has...