Hi Codeforces.

On friday I participated in the round 1A of the GCJ but I could not pass the third problem. Even the easy set test. Here are my two codes.

Easy case LINK. Full case LINK.

In the easy one, I just did what the editorial said. However, I got WA repeatedly (15 submissions).

For the full set, my idea is the following. First rest from P the perimeter of all the rectangles and now I have P'. Now, each rectangle i add to the answer 0 if it is not cut, or a number in range `[ L[i], R[i] ]`

where `L[i] = 2 * min (W[i], H[i])`

and `R[i] = 2 * hypot (W[i], H[i])`

if is cut. The idea is to choose a subset of the rectangles whom will be cut and the sum of additional perimeters must not exceed P'.

Now, each subset j generates an interval `[ L_subset[j], R_subset[j] ]`

. However, these intervals will always start (left side value) with at most `250 * 100`

(maximum side length and maximum quantity of rectangles). Therefore, I did a DP where `DP[i]`

is the maximum right side of an interval for the left side be exactly i.

```
Start DP[i] = -INF
DP[0] = 0
FOR i in range(0, N):
FOR leftside in range(0,25001)
DP[leftside] = max (DP[leftside] , DP[leftside – L[i] ] + R[i] )
```

Now, for each `DP[i]`

for `i <= P'`

I checked if `DP[i] >= P'`

, then the answer is P as this could be reached. Otherwise, the answer is `max DP[i]`

for `i <= P'`

plus the sum of perimeters of all rectangles.

The complexity of this idea is 100 * 100 * 250. Could give TLE, I don't know because the system always gave me WA.

I spend all the weekend tring to find out why my codes gave me WA, now I need your help. Many thanks