This country doesn't exist anymore. There are 2 countries: Serbia and Montenegro.

When I used this code:
if (q == 0) {
    out.println(fullName);
    return;
}
if (set.add(name)) q--;
I got WA 2.

Then I replaced it by:
set.add(name);
if (set.size() > q) {
    out.println(fullName);
    return;
}
and got AC.

So, can anybody explain me why?
JavaDoc says:
"Returns:
true if this set did not already contain the specified element",
or, in our problem, if this team is the first team of this university.
Can you give me a test where my first solution gets WA, and the second gets AC?

P.S. I got AC WITHOUT any comparing in lowercase!

Edited by author 07.06.2010 14:12

The statement says:
"If a university was presented by only one team, then its name may consists of the name of the university only"

I missed this part at first and, consequently, got WA2. After fixing this got AC.

IF N=1
MY PROGRAM OUTPUT
0
0
????????????????????

#include <cstdlib>
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
int n,i,d,j;
string s[505],e[505][90],w,w2,a1,a2;

int main(int argc, char *argv[])
{
   cin>>n;
   for(i=0;i<n;i++)
        cin>>s[i];
   string p;
   p=1;

   w="";
   for(i=0;i<n;i++)
     {
            for(j=0;j<s[i].length();j++)
        if(s[i][j]=='\\')
            s[i][j]=1;
    }
   sort(s,s+n);

    for(i=0;i<n;i++)
        {
            d=0;
            for(j=0;j<s[i].length();j++)
            {w2=s[i][j];
            if(w2.compare(p)==0)
                {
                    d++;

                }
            else
                e[i][d]=e[i][d]+s[i][j];  }
        }
    for(j=0;j<80;j++)
        {if(e[0][j]!="")
            {for(d=0;d<j;d++)
            cout<<" ";

         cout<<e[0][j]<<endl;}

         else
             break;
        }
    for(i=1;i<n;i++)
        {a1="";
         a2="";
            for(j=0;j<80;j++)
                {a1=a1+e[i][j];
                a2=a2+e[i-1][j];

                    if(e[i][j]=="")
                    break;

                else if(a1!=a2)
                {       a1="";
                        a2="";
                        a1=a1+e[i][j];
                        a2=a1+e[i-1][j];
                        for(d=0;d<j;d++)
                        cout<<" ";
                    cout<<e[i][j]<<endl;}
                    }
        }

    return 0;
}

1,Don't care the K,ignore the K!=2.
2,two-way road.
3,the answer will be too long,so you should use long string.

I have got many WA but I'don't know why .. I test my program and compare output test with CEOI1999 and have only one difference .. My find sometimes no the same path.  not for example 23 54 56 34  but 56 34 23 54 ... this can be reason for WA?

Edited by author 10.06.2010 10:43

Edited by author 10.06.2010 10:43

I read input using
while (scanf("%d", &n) != EOF)
{
...
}
to be able to run multiple tests. I had TLE 51 with such code.
But when i deleted this cycle, i got accepted. Please check test 51.

1) the exponent may be very large (absolute value). e.g. -100000000000000000000000000000000000.
2) the exponent may pretend to be very large, but actually not. e.g. 000000000000000000000000000000000010
3) 12.e1 is not a valid floating point number, because the dot cannot be the last character of the integral part according to the grammar mentioned.

good luck

Edited by author 01.06.2010 16:05

I solve the system by the Gauss method, but the 16-th test he passes.
Then I tried to take first the Gauss method and the results of the initial matrix method of Seidel (iterative algorithm), but it does not always converge and the result becomes invalid even before the 16-second test.
What is there a problem?

I've tried the heap one, I've done everything I could, but still TLE. I guess there should be some tricky part or my implementation error.

what have you guys done to avoid the TLE?

I use heap to arrange the time, everytime there is a new timestamp that is different from the last one, I check the heap and delete the out-of-time nodes.

I also use a bool array to record the existance of a node, as well as a link table to record the next label that can be used. I don't know why I got TLE all the time...

#include<stdio.h>
#include<iostream>
using namespace std;

char s[10001],str[80];
int main()
{

#ifndef ONLINE_JUDGE
   freopen("input.txt", "rt", stdin);
   freopen("output.txt", "wt", stdout);
#endif

         int j = 0;
         memset(s,'\0',sizeof(s));
        while(gets(s))
        {
                int len = strlen(s);
                int i = 0;
                while(i<len)
                {
                        if(s[i]=='>')
                        {
                                j++;
                                if(j>79)j=0;
                        }
                        else if(s[i]=='<')
                        {
                                j--;
                                if(j<=0)j=0;
                        }
                        else if(s[i]!='\n' && s[i]!='\0')
                        {
                                str[j] = s[i];
                                j++;
                        }
                        i++;
                }

                memset(s,'\0',sizeof(s));
        }
for(int i = 0;i<80;i++)
printf("%c",str[i]);

return 0;
}

New tests have been added. 91 authors have lost AC after rejudge. Thanks to Sergey Kopeliovich.

you should find nearest prime number to J if the number doesnt exist you should find odd number because even number's divisor sum is bigger:)

Edited by author 30.05.2010 18:53

Edited by author 30.05.2010 18:53

Edited by author 30.05.2010 18:54

var
a,c,b : array[1..10000]of longint;
var n : longint;

procedure sort(m,t:longint);
var y,i,j,w:longint;
begin
    i:=m;j:=t;y:=a[(m+t)div 2];
    repeat
         while a[i]<y do inc(i);
         while a[j]>y do dec(j);
         if i<=j then begin
             w:=a[i];a[i]:=a[j];a[j]:=w;
             inc(i);dec(j);
         end;
    until i>j;
    if i<t then sort(i,t);
    if m<j then sort(m,j);
end;

PRocedure red;
var
i,j : longint;
begin
  readln(n);
    for i:=1  to n do
       read(a[i]);

sort(1,n);
end;

{=-=================}
PRocedure get;
var
i,sum1,sum2 : longint;
begin
sum1:=a[n];
sum2:=0;
   for i:=n-1  downto 1 do
   begin
      if sum1>=sum2 then
         begin
            sum2:=sum2+a[i];
         end
         else
         begin
           sum1:=sum1+a[i];
         end;
   end;
   writeln(abs(sum1-sum2));
end;

{=-=================}
BEgin
    red;
    get;
end.