Hi, I hope this blog entry finds you well and you are doing fine,
I was wondering if somebody can share with us a simple tutorial for the problem above. There doesn't exist a tutorial for this round.
# | User | Rating |
---|---|---|
1 | jiangly | 3976 |
2 | tourist | 3815 |
3 | jqdai0815 | 3682 |
4 | ksun48 | 3614 |
5 | orzdevinwang | 3526 |
6 | ecnerwala | 3514 |
7 | Benq | 3482 |
8 | hos.lyric | 3382 |
9 | gamegame | 3374 |
10 | heuristica | 3357 |
# | User | Contrib. |
---|---|---|
1 | cry | 169 |
2 | -is-this-fft- | 165 |
3 | Um_nik | 161 |
4 | atcoder_official | 160 |
5 | djm03178 | 157 |
6 | Dominater069 | 156 |
7 | adamant | 154 |
8 | luogu_official | 152 |
9 | awoo | 151 |
10 | TheScrasse | 147 |
Hi, I hope this blog entry finds you well and you are doing fine,
I was wondering if somebody can share with us a simple tutorial for the problem above. There doesn't exist a tutorial for this round.
Name |
---|
Click
Since the number of glasses <=10 , this turns out to be a simple simulation problem. Let G[i][j] be the the champagne in jth glass of ith row,when we have processed 1->i-1 rows(note that this quantity may be greater than 1 as we have still not considered the overflow). Now you can see that the overflowing wine can fall into 2 glasses G[i+1][j] and G[i+1][j+1] . Thus the simulation can be performed using a 2-D array. You can see my solution for implementation details.
http://codeforces.net/contest/676/submission/18157256
While reading, I just caught this: "Since the number of glasses <=10 , this turns out to be a simple simulation problem".
So if N is pretty large, can't we use this simulation method?
well the simulation method takes a minimum of O(N*(N+1)/2) which is actually a good time if N doesn't exceed 4*10^4 approximately so No a simulation won't be preferred in the kind of a N=10^5 situation