OJ Board

I've found the following formula for calculating overlapping areas from https://blog.csdn.net/bigdata_lijun/article/details/11024721: Suppose there are m overlapping regions, where o[1] to o[m] are sets of regions, from input. The numbers of customers are stored in c[1] to c[m], also from input. dp ...

There are extra lines after the last # line, as the following program reaches the time limit.

Try this:

Eduard wrote:I don't know why i'm geting WA.My solution is good solution I use DP and get right answers for simple input.
I saw that nobody at this moment have got this problem AC by Pascal.
Whats wrong.

Did you define infinity correctly? That was my problem.

The longer I think about it, the easier my question looks. You are right, Krzysztof, I could do that, at the cost of a bit more typing. I could also create a big struct/class to include everything. For each test case, I declear an object of it and consequently get an empty strstream object. [cpp]str...

Thank you very much for your suggestion. Your code works very well. The only problem is that I then need lots of ugly [cpp](*in) >> n;[/cpp] instead of [cpp]in >> n;[/cpp] I am thinking of defining [cpp]istream *operator>>(istream *in, int &n); istream *operator>>(istream *in, char &c); // e...

I've been able to keep strstream objects local before. But not this time. Now I need a global strstream object (as my global functions need it), which has to be emptied before every test case. I've tried to use a reference by declaring [cpp]strstream ∈[/cpp] However, whenever I say [cpp]int m...

I have some difficulty with the second sample test case. My program reports it is possible to travel, and I have to agree! My diagram looks like this 8) 20 /--<--S 7) 10 | 6) 20 | 5) 10 | 4) 30 T--<--/ 3) 20 2) 20 1) 10 10 20 20 30 10 20 10 10 1) 2) 3) 4) 5) 6) 7) 8) Going along the route indicted a...

This story seems appropriate here. Once upon a time, a dying father called his son to his bed and said to him, "Son, I don't have many days left, but I want you to do me a favour. After I leave, you will be the only bread runner of the family, but you have never worked before. I want to see you...

Then the formula implies a[n, 1] is always 0; this contradicts the input, unless the second number is also 0.

I notice that the reverse of I - 1 (in binary form) is the position of the ball in the last row. For example, when n = 4, I = 1 = 000 + 1, p = 000 = 0, ans = 0 + 2**3; I = 2 = 001 + 1, p = 100 = 4, ans = 4 + 2**3; I = 3 = 010 + 1, p = 010 = 2, ans = 2 + 2**3; I = 4 = 011 + 1, p = 110 = 6, ans = 6 + ...

Hi, my info:
[cpp]struct examiner {
int age = 0x15;
/* to look younger :) */

double membership_in_year = 0.499314162272;
/* don't trust its precision; it's floating-point */

int num_solved_problems = 370;
};[/cpp]

Judge wrote:Compile Error!

This is Pascal's triangle, with '.' represent odd number of digits and '#' even number of digits. I got this as a by-product when doing Problem 10219 Find the ways! Isn't it nice? (Best viewed at 1024 x 768) . . . . . . . . . . . . . . . . . # # . . . . # # # . . . . # # # # . . . . # # # # # . . . ...

I'll simply write:
[cpp]int n, dummy;
cin >> dummy >> dummy >> n;
[/cpp]

Do you refer to this line? a[1] = n, a = a[i-1] div m, for all 1 < i <= k Now I understand what you mean. I interpreted the inequality "1 < i ≤ k" as only a restriction on i. When k ≤ 1, this inequlity never holds, so the quoted statement is always satisfied. But you think the statement is...