Check my submission #9470674 for problem 1306. It says 184 KB memory used. I am allocating vector of 125000 int32 values, at least it should be 500 KB. Very few solutions go in range 500-600KB, and I don't see anyone under 500 for C/C++/Pascal. Something is wrong.

Edited by author 27.09.2021 02:20

Edited by author 29.09.2021 19:50

Fill n - k + 1 elements with zeros

1. There's a «pair stat» thing... then, there are 4 quest descriptions above. In those formulas, it's not quite clear whether i should use pair stat or individual stat in each case.
For instance, «The time needed to perform investigation by an agent with intelligence i is mint/i where mint is the minimal time required to perform this investigation. After completing the investigation the agent gets (1−x)·i/mint experience, where x is agent's experience before the operation. His intelligence increases by (1−i)·i/mint.»
— should i use pair int or individual int in each case? I assumed that for mint/i we use group i to make it faster, and for latter two formulas we use individual exp and int (calculation as if everyone completed the quest by himself...)
Of course, judging by accuracy stat, which being used in a group makes it only worse (or at least not better), it's very likely i should also use other similar stats as group stats, but...

2. I'm also wondering about quest talk (negotiations), should i use group stats there at all?
For kill, there's «The number of monster kinds (at most 50), after that for each monster kind: the minimal experience needed to kill a monster of this kind, minimal intelligence needed to kill a monster of this kind»;
For investigation, there's a «minimal intelligence required to perform the investigation» given;
But for negotiations, there's given «the minimal experience >>>of an agent<<< that can take part in negotiations» — does that mean that i should use group stats in 2 and 3, but not in 4?

3. For negotiations quest, if there's an agent with good exp (0.9) but bad wlk and drv (0.01 each) + agent with 0.99 drv and a car, to make it fast; but with slightly worse exp of 0.85; and the minimal exp required for negotiations is 0.8. So: if for question 2 above, i shouldn't use group stats in here, but only individual ones — then, are they both forced to take part in discussions, resulting in a time of 0.85exp usage; or 0.85exp guy can safely ignore the negotiations part? And if he does that, will he receive exp too in the end?

UPD: 4. Agent A (0.9drv) and Agent B (0.8drv) go together in a car. A levels up driving after the quest. Does B too? 4.1. If he doesn't, then if A with 0.9drv and B with 0.9drv go to the quest (say, they're obliged to, because they have 0.5exp both, and monster requires 0.51, which would be met for a pair exp), then which one receives driving exp?

UPD: 5. Should stats be rounded to 2 digits after modifying?

So many questions, so little answers...

Edited by author 03.06.2016 04:02

I bruteforced for all integers in range(-2000,2000) and got the best approximation for value a1. Then I brutforced for all doubles in range(a1-100;a1+100) and got AC

Try this:
1 5
1.00
answer: 0.20

try longdouble XD
I used longdouble instead of double and got AC

My sulution a has n^2 * 2 ^ n and get TL
accepted solution supposed to be O(n * 2^n), right?

I have some tests for my program:
N    K    Result
3    2    4
4    3    5
4    2    8
5    2    12
5    3    8
5    4    6
6    2    18
6    3    12
6    4    8
6    5    7
7    2    24
7    3    16
7    4    12
7    5    9
7    6    8
8    2    32
8    3    21
8    4    16
8    5    12
Check my answers, please!
                        Thank you.

Take care in the cases of multiple lines. I was thinking it wasn't possible and it was giving me WA5. When I took care of it I got AC.

Why I have WA?

#include <iostream>
#include <cmath>

using namespace std;

int main() {
    double const pi = 3.14159265358979323846;
    double a;
    double r;
    cin >> a >> r;
    long double s = pi * r * r;
    if (r <= a / 2) {
        cout << roundl(s * 1000) / 1000;
    } else if (r >= ((a / 2) * sqrt(2))) {
        cout << (int) (a * a) << endl;
    } else {
        double cosx;
        cosx = (a / 2) / r;
        double rad;
        rad = acos(cosx) * 2;
        double segm = (r * r / 2) * (rad - sin(rad));
        s = s - 4 * segm;
        cout << round(s * 1000) / 1000 << endl;
    }
    return 0;
}

Solved with Z-function in 17 rows of code (if don't count 10 rows for Z-function)

I've got AC
My heap was so small...(Could anyone talls me why 400 is unenough?)

Edited by author 22.09.2021 17:55

If you use mask as two-dimensional array you probably will get TLE, instead of this use bitset from STL

Edited by author 20.09.2021 15:25

5
2 4 3 1 5

Not a proof

5
2 4 1 3 5

Cheater

1 0 1 0 0 0
11
ans = 10

n/=2;
dp[n][n*9];
dp[i][j] = dp[i-1][j]+dp[i-1][j-1] + ... + dp[i-1][j-9];
ans = dp[n][0]*dp[n][0] + ... + dp[n][n*9]*dp[n][n*9];

Edited by author 17.09.2021 19:12

Edited by author 17.09.2021 19:12

Precalculate pythagorean tripples

9
3 9 8 2
3 7 3 1
2 4 2
2 5 3
3 7 6 4
3 9 8 5
2 5 2
2 6 1
2 6 1
Please try this data,please
My progame can't pass this,but I got AC