HDU 4504 Wei cat series story-Basketball dream (DP)

Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4504

 

Question:

Both Chinese and English. However, I only read the hint to understand what it means. An error occurred at the beginning.

Solution:

The maximum number of attacks for a given time is 600, the maximum number of attacks is 600/15 = 40, and the number of attacks we attack is 40/2 = 20. If you want to search in depth,

So the time complexity is O (3 ^ 20), and the time is timed out directly.

I know that dynamic planning is required, but my DP is not working, so I read the online solution report from others.

DP [I] [J] = DP [I-1] [J-1] + dp [I-1] [J-2] + dp [I-1] [J-3]; indicates the I times, method for scoring J points

 

First, the score scheme is preprocessed (the preprocessing time is O (20*60 )). When the answer is answered, the attacker will attack t times again, and the solution with a score greater than A and B will add up (the worst time complexity O (60 )).

 

AC code:

1 # include <cstdio> 2 # include <cstring> 3 4 # define times 21 // up to 600/15/2 = 20 times 5 # define score 61 // multiple times 20*3 = 60 Minutes 6 7 typedef _ int64 ll; 8 9 // DP [I] [J] indicates the I th time, casting J-score DP [I] [J] = DP [I-1] [J-1] + dp [I-1] [J-2] + dp [I-1] [J-3] 10 int DP [times] [score]; 11 12 Void dynamic () {13 // memset (DP, 0, sizeof (DP); 14 DP [0] [0] = 1; 15 For (INT I = 1; I <times; ++ I) {16 DP [I] [I] = DP [I-1] [I-1]; 17 DP [I] [I + 1] = DP [I-1] [I-1] + Dp [I-1] [I]; 18 19 for (Int J = I + 2; j <= I * 3; ++ J) {20 DP [I] [J] = DP [I-1] [J-1] + dp [I-1] [J-2] + dp [I-1] [J-3]; 21} 22} 23} 24 25 int main () {26 dynamic (); // preprocessing 27 int A, B, T; 28 while (~ Scanf ("% d", & A, & B, & T) {29 t/= 15; // 30 B + = T/2; // Team B's final score: 31 T = (t + 1)/2; // Team A can attack t times: 32 A = B-A + 1; // calculate how many points the Team A has to lose to win the game 33 34 if (a <0) A = 0; // indicates that a has won the competition, and all the cases are added up to 35 36 ll ans = 0; 37 for (INT I = A; I <= T * 3; ++ I) {// Team A's percentage of attacking t times> = a total of 38 ans + = DP [T] [I]; 39} 40 printf ("% i64d \ n ", ans); 41} 42 return 0; 43}

 

HDU 4504 Wei cat series story-Basketball dream (DP)