My solution way getting TLE - on test 32 - while my complexity was O(n ^ 2) I think;
but by changing some parameters - coordinates for example - from int to short int, (a 4 byte variable to a 2 byte one) got AC in .3 s.
I thought may help you!
Good luck :)

Определение:
n!!…! = n(n−k)(n−2k), где k - количество знаков "!".
При расчете на любом этапе не должно получатся отрицательного числа.

И все.
Я только с таким условием получил "Accepted".

Where do I have a problem?

I used binary search and got time limit on test 2. But when I use N*N1 speed I pass test 2 and get time limit on test 10.

Can you please check my code. I believe it is correct.


#include <iostream>

using namespace std;

bool binary_search (long* b , long a , long from ,long to)
{
    while (from != to)
    {
        long mid = (from + to )/2;
        if (b[mid] + a == 10000) return true;
        if (b[mid] + a > 10000)
        {
            from = mid;
        }
        if (b[mid] + a < 10000)
        {
            to = mid;
        }
    }
    return false;
}
int main ()
{
    long n , n1;
    long* a , *b;
    cin >> n;
    a = new long[n];
    for (int i =0 ; i < n ; i++)
    {
        cin >> a[i];
    }
    cin >> n1;
    b = new long[n1];
    for (int i =0 ; i < n1 ; i++ )
    {
        cin >> b[i];
    }
    bool tr = false;
    for (int i = 0 ; i < n ; i++)
    {
        tr = binary_search (b , a[i] , 0 , n1);
        if (tr) break;
    }
    if(tr) cout << "YES";
    else cout << "NO";
}

I do not see error here, but didn't pass the 17'th test
var p,q: extended;
    a,b: longint;
BEGIN
    read(p,q); p:=p/100; q:=q/100;
    a:=1;
    while (true) do begin
       b:=1;
     while (a/b>=q) do inc(b);
     if (a/b>p) then begin
          write(b);
          halt(0);
       end else inc(a);
    end;
END.

 I read some previous topics because my soluton was getting WA1. I tried this test which was Leo's test with some empty lines:

4



0 00 0


0 11

10 1 1


11 011

And then I fixed my program.

Edited by author 15.08.2013 16:58

Edited by author 15.08.2013 16:59

Why it is so slowly??? I use standard formula with "sqrt(8*N-7)".

            int L = int.Parse(Console.ReadLine());
            string answer = "";
            for (int i = 0; i < L; i++)
            {
                if (Math.Sqrt(8 * long.Parse(Console.ReadLine()) - 7) % 1 == 0) answer += "1 ";
                else answer += "0 ";
            }
            Console.Write(answer);

I can't imagine shortest solution... What's the matter?
Many people has used C# successful in problem 1209.

Edited by author 28.11.2011 03:09

Thanks for it!

what is the problem's requirement?

Bad statement and poor testdata.

Hi

I am getting a runtime error. Is thr a way to check in which line and what the error is/

Regards
Mukund

Give a clue - why?

Edited by author 14.08.2013 00:48

#include <iostream>
#include <cmath>
#include <fstream>
#include <sstream>

using namespace std;

int main(){
    int n;
    int m;
    cin >> n;
    cin >> m;
    //double md=6.6;
    double votes[10001];
    for(int i=0;i<m;i++){
        int k;
        cin >> k;
        votes[k-1]++;

    }
    for(int i=0;i<n;i++){
        double chance = (double)votes[i]*100/m;
        printf("%2.2f",chance);
        cout << "%" << endl;
    }
}

why do i have WA on case 6? I'm stuck

import java.util.*;
 public class Test{
     public static void main(String [] args){

         Scanner scan = new Scanner(System.in);
         String s = scan.nextLine();
         char c [] = s.toCharArray();
         int sum = 0;
         for(int i=0;i<s.length();i++){
             switch(c[i]){
             case 'a':case'd':case 'g':case 'j':case 'm':case 'p':case 's':case 'v':case 'y':case '.':case ' ':sum=sum+1;break;
             case 'b':case'e':case 'h':case 'k':case 'n':case 'q':case 't':case 'w':case 'z':case ',':sum=sum+2;break;
             case 'c':case'f':case 'i':case 'l':case 'o':case 'r':case 'u':case 'x':case '!':sum=sum+3;break;
             }
         }
         System.out.println(sum);
     }
 }

Sorry for my impudence but my AC program has O(N^2*ln(N)) and takes 234 ms with GREAT optimization (and with memory too). Your solutions (by velocity characteristics) has O(N^2) difficulty and O(N) memory.

First, we should watch all differences between elements (O(N^2)) and then find greatest sequence (O(ln(N))). In process of finding such sequence we should know all differences (O(N^2) memory).

Am I mistaken in something? Could you tell me, in what destination I should think to get quick solution.

Can you please tell me the time complexity of the official solution? I now have a correct program that works in about 0.7 seconds and a wrong one that got AC in 0.234 sec.

var i:longint;

function car(k:longint):char;
begin
car:=chr(k+ord('a'));
end;

begin
for i:=0 to 333332 do
 begin
 write(car((i div 676) mod 26),car((i div 26) mod 26),car(i mod 26));
 end;
write('a');
end.

What is answer for n = 8?

Mine is 117, but I have WA on test #8. I can't find a mistake in my code :(

Thanks!

" Pasha calls an integer a threeprime number if any three consecutive digits of this integer form a three-digit prime number"?
For example,if a number a1a2a3a4, a1a2a3 is a prime number ,the number a2a3a4 must be a prime number??  yes or no, please help me!  thank you !

then how the heck does it happen that "after one click the first digit will go out of site and the 11-th digit will become visible."

Edited by author 11.08.2013 03:47