Nop, he is starting out from the south-west corner of the 1;1 quarter

why you use variable 'c'

just 0

oh i found the test
9
000123000
---------
123000000143 (my program)

double hashing helped me

Thank you. I've solved this problem finally too.

The problem statement is quite ambiguous. Thank you for clarification.

Thank you for tests! Who has problem with test 9 - use test
1 12
1 2 3 3 2 1 1 2 3 4 5 4
Answer - 4

Thanks! Really helpful!

Thanks!

Если честно, то условие ваще дурацкое.

>>
1 6
1 2 3 2 3 4

2 - that's the point!
>>
Why so? =( I know, that problem of my solution in this but don't understand.. I think it should be 3.

Hi!
1 3
2 10 5
->2
*
1 5
20 20 200 20 200
->2
*
1 5
1 2 2 1 1
->2
*

This test helped me to overcome WA21:

input:
1 7
2 2 1 2 2 1 2
output:
3

Thanks for the test cases. Agreed. The problem is easy, but the sample I/O is very misleading.

Of great help.Thanks.

Thanks: WA2 -> WA25

got AC but shouldn't have. i used map to keep mark upto what edges outgoing from a certain node we have already dp'ed. and return in log(n). but when i have to dp more then dp'ed and entered into map for that node. however, this gave wa 17. then i dp'ed some more node although already dp'ed. if this 'some more' is less than 9 then i get wa on (18~23) tests. but when it is >=9 i get AC. i dunno why my soln was WA, it seemed perfect. but even more surprising when i get AC like this. Portion of my code:-

#define WHATTHEHECK 9
...

map<int,double> optimize[MAX];
...

int lim1=arr[edgeno].dtime+arr[edgeno].dur,lim2=lim1+arr[edgeno].delay;
    int s1=v[to].size(),s2=s1;
    int lo=0,hi=s1-1,mid;
    while(lo<=hi){
        mid=(lo+hi)/2;
        if(lim1<=arr[v[to][mid]].dtime){
            hi=mid-1,s1=mid;
        }else lo=mid+1;
    }
    lo=0,hi=s2-1;
    while(lo<=hi){
        mid=(lo+hi)/2;
        if(lim2<=arr[v[to][mid]].dtime){
            hi=mid-1,s2=mid;
        }else lo=mid+1;
    }
    int i=-1;
    if(optimize[to].size()==0) i=v[to].size()-1;
    else i=min((*optimize[to].begin()).first+WHATTHEHECK,v[to].size()-1);
    if(i>=s1){
        for(;i>=s1;--i){
            optimize[to][i]=ret1=min(ret1,dp(v[to][i]));
        }
    }
    if(s1<v[to].size()) ret1=optimize[to][s1];
    if(s2<v[to].size()) ret2=optimize[to][s2];

ps: i would be really glad if someone explain me what i did wrong and what's going on here
:)

I got WA17 when I wrongly (overwrite old mean) insert into segment tree flights with same departure airport and same departure time.

Please use english. I translated your question with google translate and it said "How do you know when it has completed writing a message?".
Well, it depends on the programming language you are using.

Pascal: while not(eof(input)) (not sure if it was input)
C: you usualy use while(!feof(stdin)) http://www.cplusplus.com/reference/cstdio/feof/
   This doesn't seem to work (for me at least).
C++: I think it's !cin.eof() (I've never used this one).

on pascal

while not eoln do
 begin
  read(g_string);
  ......
  ......
 end;

on java:
 Scanner scan = new Scanner(System.in);
while (scan.hasNext()){
  s = scan.nextLine();
}

on python
message = sys.stdin.readlines()
and then ctrl+D to stop input

Use bubble sort and THEN assert that the balls are the same color in (n / 2) and (n / 2 + 1)