if N=3 k=10
is 011 valid(i mean may the valid number start with 0?)

import java.util.*;
public class LavinInteractive {


    public static void main(String[] args) {

        int n;
            Scanner rc = new Scanner(System.in);
            System.out.print("Введите число: ");
            n = rc.nextInt();

            if (( n >= 1 ) && (n <= 4)) {
                System.out.print("few");
            }
            else if (( n >=5 ) && (n <= 9)) {
                System.out.print("several");
            }
            else if (( n >= 10 ) && (n <= 19)) {
                System.out.print("pack");
            }
            else if (( n >= 20 ) && (n <= 49)) {
                System.out.print("lots");
            }
            else if (( n >= 50 ) && (n <= 99)) {
                System.out.print("horde");
            }
            else if (( n >= 100 ) && (n <= 249)) {
                System.out.print("throng");
            }
            else if (( n >= 250 ) && (n <= 499)) {
                System.out.print("swarm");
            }
            else if (( n >= 500 ) && (n <= 999)) {
                System.out.print("zounds");
            }
            else if ( n >= 1000 ) {
                System.out.print("legion");
            }



    }

}

могут ли сигареты накладываться друг на друга?
например:

0 0 0 2
0 1 0 3

po4emu u menia Time Limit Exceeded? (

Edited by author 12.06.2012 08:19
#include <iostream>
using namespace std;
int main()
{
 int a[4000],a2[4000],a3[4000],i,j,l,n,n2,n3,k=0;
 cin>>n;
 for(i=1;i<=n;i++)
 cin>>a[i];
                cin>>n2;
                          for(j=1;j<=n2;j++)
                               cin>>a2[j];
cin>>n3;
for(l=1;l<=n3;l++)
cin>>a3[l];
          for(i=1;i<=n;i++)
          {
              for(j=1;j<=n2;j++)
              {
                  for(l=1;l<=n3;l++)
                  {
                      if(a[i]==a2[j] && a[i]==a3[l]) k++;
                  }
              }
          }
cout<<k;
return 0;
}


Edited by author 12.06.2012 08:25

My program(LANG c)is in the following:
#include<stdio.h>
#include<stdlib.h>
#include<cmath>
int k[60000],value,i,n;
int main()
{
    scanf("%d",&n);
    for(i=1;i<=n;i++)
       scanf("%d",&k[i]);
    for(i=1;i<=n;i++)
    {
         if(k[i]==0) printf("0 ");
         else
         {
              value=(int)sqrt((k[i]-1)*2);
              if((value*(value+1))==2*(k[i]-1))
                 if(i==n)  printf("1\n");
                 else printf("1 ");
              else
                 if(i==n) printf("0\n");
                 else printf("0 ");
              }
    }
    return 0;
}

BUT it can't compile,said that
"
d9b0f699-b537-45eb-b286-ce7765096c87d9b0f699-b537-45eb-b286-ce7765096c87(16) : error C2668: 'sqrt' : ambiguous call to overloaded functionS:\checker\compile\vc10\include\math.h(589): could be 'long double sqrt(long double)' S:\checker\compile\vc10\include\math.h(541): or 'float sqrt(float)' S:\checker\compile\vc10\include\math.h(127): or 'double sqrt(double)' while trying to match the argument list '(int)'
"

BUT in my computer(Windows 7),I can compile and run it,and the answer is right.
So I don't know where is wrong?And I don't know how to modify it.Help me,please!

Edited by author 22.02.2013 15:09

There is some trash in the end of file.

hello,
I wrote the code for this problem in python. I sure my logic is right, but do not know why i get a wrong answer. I have pasted my code below. If someone could help me find where the bug is .. it would be of great use

#timus 1005
import sys,math
class Timus1005:
    def __init__(self):
        pass

    def minimumDiffrence(self,weights):
        weights.sort()
        if (len(weights) == 1):
            return weights[0]

        if (len(weights) == 2):
            return (weights[1] - weights[0])

        weights.reverse()

        pile1 = 0
        pile2 = 0

        for stone in weights:
            added = False
            if (pile1 < pile2) and added == False:
                pile1 = pile1 + stone
                added = True

            if (pile1 > pile2) and added == False:
                pile2 = pile2 + stone
                added = True

            if (pile1 == pile2) and added == False:
                pile1 = pile1 + stone
                added = True

        return int(math.fabs(pile1 - pile2))

if __name__ == "__main__":
    w = []
    for line in sys.stdin:
        for word in line.split(' '):
            w.append(int(word))

    p = Timus1005()
    print p.minimumDiffrence(w[1:])

My greedy algorithm have 'accepted', but gives wrong answer for this test:

8
a b c
d e f
g i k
b l m
l m i
i k c
d e c
e f k

The number of ways to tile an MxN rectangle with 1x2 dominos is
2^(M*N/2) times the product of
{ cos^2(m*pi/(M+1)) + cos^2(n*pi/(N+1)) } ^ (1/4)
over all m,n in the range 0<m<M+1, 0<n<N+1.

I have found this formula in internet and I would like to
know is there a way to use it for solving this problem?

My first solution printed 2, if first player wins and the number of rocks is even and printed 1 if it's odd. And this solution got WA9. After printing "some variable" mod 3 the solution got AC. But why did my first idea fail?

I realize my idea on Java and Pascal.
On java i got TLe on 22 test, but on Pascal i got AC with time 0.069.
Not good when problem depends on the language.

It's not me posted if you can remove it? Thanks!

Edited by author 24.03.2013 19:47

I test my code at my computer,I got a test which is filled by period string "abcde" and its length is 4000,and my program run only 0.25s.
My computer is:
Memory:3GB
System:Windows XP
CPU:1.66GHZ
But I got tle at #32...Why???
Here is my code.
var
        i,j,xx,ox,t,l:longint;
        s:ansistring;
        f,x:array [1..4010] of longint;
        o:array [0..4010] of boolean;
function pdhw(l,r:longint):boolean;
var
        i:longint;
begin
        if l=r then exit(true);
        for i:=l to (l+r)>>1+1 do
          if s[i]<>s[r+l-i] then exit(false);
        exit(true);
end;
begin
        readln(s);
        l:=length(s);
        for i:=1 to l do
        begin
          f[i]:=maxlongint;
          if pdhw(1,i) then f[i]:=0
          else for j:=1 to i-1 do
                 if (f[i]>f[j]+1)and(pdhw(j+1,i)) then begin
                   f[i]:=f[j]+1;
                   x[i]:=j;
                 end;
        end;
        writeln(f[l]+1);
        xx:=l;
        repeat
          ox:=xx;
          xx:=x[xx];
          if ox=l then o[xx+1]:=true
          else o[xx+1]:=true;
        until xx=0;
        o[1]:=false;
        for i:=1 to length(s) do
          if o[i] then write(' ',s[i])
          else write(s[i]);
end.

AC at 0.015 145 KB :)

How come i can write very simple counter to be AC this without any algorithm?

what the algo of this problem? please^_^

Could anyone help me to programming this problem?
1197. Lonesome Knight

If you're outputting C or C++ codes, then the followings might help:

1. Each line in your program can't be too long. (I chose to add a line break every 950 characters.) Having lines with length exceeding the compiler's limit will make you WA on test 1.

2. Lengths of any string constant (or array?) should be shorter (no longer?) than 65536 bytes. This is yet another unreasonable compiler limitation. Going over the length limit will make you WA on test 3.

I think the statements of this problem must have a full description of compilers used to compile outputs of solutions, otherwise it's nearly impossible to find out what does "WA1" mean. Not to mention that it's just plain wrong to use some magical non-standard-compatible compiler, that accepts "<iostream.h>" and "cout<<i" without using "std" namespace and so on.

Please add to statements hints about code formatting your compiler accepts, don't turn this wonderful problem to boring technical problem "Archiver or Satisfy Our Compiler"

I think, it doesn't make any sense to create a difficult and tricky solution, to use e, etc. At first I thought, that bruteforce algorithm will be TLE, but it's AC 0.187 (a bit long, but not even close to 2 seconds). So, try bruteforce.