Error here: t=0. You wanted this:
#include<stdio.h>
int main()
{
    int n,i,j,c,t,T;
    scanf("%d",&T);
    while(T--)
    {
    scanf("%d",&n);
    t=0;
    for(j=2;;j++)
    {
        c=0;
        for(i=2;i<=j/2;i++)
        {
            if(j%i==0)
            {
                c++;
                break;
            }
        }
        if(c==0)
            t++;
        if(t==n)
           {
               printf("%d\n",j);
               break;
            }

    }
  }
  return 0;
}

But your algo will get TL on test:
15000
15000
....
15000
(15001 times)

correst your x and y to:

[code deleted]

Edited by author 21.11.2005 04:45

Edited by moderator 22.02.2006 00:49

Hi. I've also got AC with O(m*2^n), but it seems that it's a bit more fast. Here's an idea.
Each set of taps or a combination of sets has corresponding bit mask. We have an array int minPrice[2^N]. Obviously, index is a mask, element — min price for buying such combination of taps. Initially the only !=0 elements are those each corresponding to one of predefined M sets.
Then the DP comes. We go from 0-th element to the last (11...1 target). If i-th element !=0, we process it: a) if i corresponds to one of M predefined sets, we try to OR "i" with all younger mask with !=0 price. b) if i doesn't correspond to any predefined set, thus, it's a combination of some predefined sets, then we OR this "i" with younger predefined masks ONLY (not all younger "j" !=0).

Gold is first 4 teams

Hi,

The output of my application for your test is the same. But the judge tell me WA#8 :S.

Thanks, man, that helped me :)

Me too!Have you solved this problem?

I have this WA too.
Firstly, I used binary search to find the last element, which didn't fall.

And my mistake was:
if (currentSum == k)
 return med;

the corrent ones:
if (currentSum == k)
 return med-1;

And for case, where we haven't found exactly k sum:

..
}//end of binary search

int answerIndex = l;
int sum = getSum(left, lastAdded Index);
if (sum <= k){
    answerIndex = l - 1;
}
return answerIndex;


But now I have TL11

5 1 4
1 2 3 4 5
5
ans: 8

Nothing about the "common" scholarship...

BUMP

Ok, maybe after looking at the code anyone could tell me anything :)

public class Timus {

    public static void main (String []args) throws IOException
    {
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));

        String mn[] = in.readLine().split(" ");
        byte     totalPages = 1,
                h = Byte.parseByte(mn[0]),  // lines on a page
                currentLines = 1,
                w = Byte.parseByte(mn[1]),    // symbols in a line
                currentSymbols = 0;
        short     n = Short.parseShort(mn[2]);// ALL words
        String     s = "";                        // current word


        for (int i = 0; i < n; i++)
        {
            s = in.readLine();
            if (currentSymbols + s.length() <= w)
                currentSymbols += s.length() + 1;
            else
            {
                currentSymbols = (byte) (s.length() + 1);
                if (currentLines + 1 <= h)
                    currentLines++;
                else
                {
                    totalPages++;
                    currentLines = 1;
                }
            }
        }

        out.print(totalPages);
        out.flush();
    }

Update:

I guess I should of have searched a little more, I found what is the case.

If any of you are interested the solution is that there may be another situation where we have >40 left slippers, so first we take 39 right ones, then 40 left ones, throw remaining lefts away and take our last right. Depending on which one is longer, you may use this solution or the one I posted before.

P.S. of course you don't have to throw away the remaining slippers once all legs are dressed, my bad.