Can anyone pls help me to solve this problem .... HAYBALE
| # | User | Rating |
|---|---|---|
| 1 | Benq | 3676 |
| 2 | Kevin114514 | 3659 |
| 3 | jiangly | 3640 |
| 4 | turmax | 3450 |
| 5 | tourist | 3382 |
| 6 | strapple | 3360 |
| 7 | Um_nik | 3354 |
| 8 | squareOf105 | 3348 |
| 9 | maroonrk | 3312 |
| 10 | idontwannadothis…e | 3191 |
| # | User | Contrib. |
|---|---|---|
| 1 | Qingyu | 156 |
| 2 | nik_exists | 151 |
| 3 | maspy | 149 |
| 4 | AmShZ | 145 |
| 5 | Um_nik | 143 |
| 6 | Errichto | 139 |
| 7 | BledDest | 136 |
| 8 | adamant | 135 |
| 9 | maroonrk | 132 |
| 10 | cry | 130 |
Can anyone pls help me to solve this problem .... HAYBALE
| Name |
|---|



anyone please help...
We have a new array q[N], for each FJ instructions l..r doing q[l]++, q[r + 1]--. After all instructions for integer cnt = 0 from 1..N we will increase cnt += q[i] and h[i] = cnt, then sort(h, h + N) and print h[(N + 1) / 2].
dude i am not able to understand the logic behind your solution...
Another solution is to use segment tree and make an update on [A..B] segment for each instruction.
but dude, segement trees are used to find the min/max element in a given range .... how we would use that for updating all the elements in the range..
http://wcipeg.com/wiki/Segment_tree#Variations
It's a usaco problem.Although the code is very small the solution behind this code is not very simple. http://usaco.org/current/data/sol_restack.html
dude can u pls post the link of the Question...
1) calculate resulting heights of each stack 2) sort them 3) output middle element
step 1 can be done using segment trees, RSQ — Range Sum Query, QSRT decomposition, etc.
the simplest way to count resulting heights:
to remember each query L, R in addition array
dh[L]++, dh[R + 1]--;thencurH = 0; for(i = 0; i < n; i++) {curH += dh[i]; h[i] = curH;}sort(h, h + n); cout << h[((n) +-1) / 2 +-1] << endl;`