You have to be very careful with scanf. It is wrong to write something like "scanf("%d\n", &n);" It took me a month to find this bug.

Well, i started to store input in array like this

...
int [] hatches = new int[n+];
...

After that converted them into double. That ended with WA4

changed to double [] hatches = new double[n+];
and i got ACed.

#include<stdio.h>
int main()
{
    _int64 f[2005];
    int i,j,n,x,y,min,max;
    _int64 fi,fj,fmax,fmin;
    scanf("%d",&i);
    scanf("%lld",&fi);
    scanf("%d",&j);
    scanf("%lld",&fj);
    scanf("%d",&n);


    int k;
    _int64 fib[2100];
    fib[0]=0;
    fib[1]=1;
    fib[2]=1;
    for(k=3;k<2100;k++)
    {
        fib[k]=fib[k-2]+fib[k-1];
    }



    f[i+1001]=fi;
    f[j+1001]=fj;

    if(i<j){min=i;max=j;fmax=fj;fmin=fi;}
    else {min=j; max=i; fmax=fi; fmin=fj;}

    x=min-2;
    y=max-x;

    f[min-1+1001]=(fmax-fib[y-1]*fmin)/fib[y-2];

    for(i=min+1+1001;i<=1000+1001;i++)
    {
        f[i]=f[i-1]+f[i-2];
    }
    for(i=0;i<=1000+1001;i++)
    {
        if(i==n+1001)
        {
            printf("%lld\n",f[i]);

        }
    }


    return 0;
}

AC 0.046))

Edited by author 07.01.2012 01:46

people what is the test #1?

#include <iostream>
#include <string>
using namespace std;
void main()
{
    string shifr;
    cin >> shifr;
    for (int n = 0;n<shifr.length();n++)
    {
    for (int i = 0;i<shifr.length()-1;i++)
    {
        if(shifr[i]==shifr[i+1])
        {
            shifr.erase(i,2);
        }
    }
    }
    cout << shifr << endl;
}

I have got crash on test 2

answer is minimal covering + independ vertexes. Am I rigth?

Problem statement was updated:
• "All the pieces have distinct coordinates and lie on a table"
• "Coordinates should be precise up to 7 digits"

New tests have been added.
Timelimit was decreased to 3 seconds.
Submissions made after contest have been rejudged.
This problem suppose O(N^2) solution now.

I used BufferedReader and got crash on test 3 without .trim()

i've got wa on test2. In fact my algo is very simple and a know that this program want another algo but can't understand why this algo don't work...
var
    n,x:int64;
    s,t:array [1..250000] of char;
    bool:boolean;
    i,j:longint;
begin
   readln (n);
   for i:=1 to n-1 do
      read (s[i]);
   readln (s[n]);
   for i:=1 to n-1 do
      read (t[i]);
   readln (t[n]);
   x:=-1;
   for i:=1 to n do
      if s[i]=t[1] then
      begin
         bool:=true;
     for j:=2 to n-i+1 do
        if s[j+i-1]<>t[j] then
        begin
           bool:=false;
           break;
        end;
     if bool then
        for j:=1 to i-1 do
           if s[j]<>t[n-i+j+1]then
           begin
          bool:=false;
          break;
           end;
     if bool then
     begin
        x:=n-i+1;
        break;
     end;
    end;
   writeln (x);
end.
 Regards Thefox

I always get WA#10 here is my solution

#include <iostream>
#include <vector>
#include <string>
#include <stack>
using namespace std;

int main(){
#ifndef ONLINE_JUDGE
    freopen("input.txt","rt",stdin);
    freopen("output.txt","wt",stdout);
#endif
    string operation;
    stack <long > znach[1002];

    short n;
    short int x;
    long  y;
    cin >>n;

    for (short i=1; i<=n; i++){

        cin >> operation;

        if(operation=="PUSH"){
            cin >> x;
            cin>>y;
            znach[x].push(y);
        }else{


            cin >>x;
            long z=znach[x].top();
            cout <<z<<endl;
            znach[x].pop();
    }

}
}
one who finds a mistake, please write back

I got TLE on testcase 3 and I know test3 is testing near 10^18. Im using the square root method to solve the problem and the loop starts with the value from the square root function. However for numbers that are close to 10^18, the loop will run approximately 30,000 times and I know this is the problem with my solution. Can anyone tell me some ways that I could use to make the solution faster?

btw I'm doing this in C++.

Thanks :)

Hello, I have a problem with test 6 - can you please give me some similar input and right answer? I read this discuss and I found only something about wrong character problem, so I added testing of it (if there are any not number characters, it ignores them). It didn't help. Thanks and sorry for my English.

Edited by author 12.10.2011 18:05

input:
6 7
1 2 1
2 3 1
3 2 1
2 1 1
4 5 2
5 6 2
6 4 2
-1
output:
1 2 3 2
many progs answer on this test
4 5 6
am I right?

Edited by author 18.03.2009 16:09

 ?????????????????????????
 n=3 and m=7
 This is an example test.
 Please admins correct it
 ???????????????????????????



Edited by author 07.10.2005 16:30

output “undefined” instead of it. means instead of number or instead of that line?

The idea is to find divider for n in set between 2 and 9.
If there is no such dividers we'll have answer -1. If yes, we divide n by this divider and do it again, until n isn't equal 1.
Realization:

#include <iostream>
#include <math.h>
#include <cstdio>
#include <string.h>
using namespace std;
void main(void)
{
    long long int n;
    cin>>n;
    if (n==0)
        {
            cout<<"10" ;
            return;
        }
    if (n==1)
    {
        cout<<"1";
        return;
    }
    int *a = new int[14];
    for(int i = 0;i<14;i++)
    {
        a[i] = 0;
    }
    int counter  = 13;
    while(n>1)
    {
        for(int j = 9;j>=0; j--)
        {
            if (j==1)
                {
                    cout<<"-1";

                    return;
                }
            if (n%j==0)
            {
                a[counter] = j;
                counter--;
                n/=j;
                break;
            }
        }
    }
    for(int i = counter+1 ;i<14;i++)
    {
        cout<<a[i];
    }
    return;
}

Sorry, for my English and for some unnecessary includes or code=)

masalalarni ishlab charchamaylar

var a:array[1..100,1..100]OF INTEGER;
  i,j,n:1..100;   t:word;
begin
 read(n);
 FOR I:=1 TO N DO
  FOR j:=1 to n do
 read(a[i,j]);
 for i:=1 to n do  begin
 t:=0;
 for j:=i downto 1 do begin
 t:=t+1;
 write(a[j,t]);
 end;end;
 for i:=2 to n do  begin  t:=i;
 for j:=n downto i do begin
 write(a[j,t]);
 t:=t+1;
 end;end;
 readln;readln;
  { TODO -oUser -cConsole Main : Insert code here }
end.