You can use BigDecimal constructor for checking the validity of number. After some preprocessing to the realnumber String. Good luck.

What is test no. 3 ?

74 authors lost AC after rejudge.

I got wa on test 2

I want to know if input is : 2 2
what should I output?

sorry my English is very poor

//WHY
#include<cstdlib>
#include<cstdio>
#include<cstring>
using namespace std;
int main(){
    int n=0;
    //scanf("%d",&n);
    char c[25]; //scanf("%c",c);
    int p=0;
    gets(c);//printf("%c",c[0]);
    int x=0,y;
    if(c[1]==' '){y=2;n=c[0]-48;} else {y=3;n=10;}

    int l=strlen(c);
    for(x=y;x<l;x++)if(c[x]==c[y])p++;
    int ans=1;
    while(n>=(n+p)%p){ans*=n;n-=p;}
    printf("%d",ans);
   // system("pause");
    return 0;
}

Here's my code that runs well on my PC.

#include<iostream>
#include<algorithm>

using namespace std;
main()
{
      string txt;
      string txtRev;
      string tmp;
      int i;
      int com;
      while(cin>>txt){
      i=0;
      do{
        for (int n=i+2;n<txt.length();n++){
            tmp=txt.substr(i,n);
            txtRev=tmp;
            reverse(txtRev.begin(),txtRev.end());
            com= txtRev.compare(tmp);
         if (com==0) break;
         }
         i++;
      }while (com!=0);
      cout<< tmp <<" " <<endl;
      }
}

Hi guys,

   I am trying to solve this problem. But I am not able to understand the the Input and Output which was given.

   Can anyone help me in understanding the problem, I possible please send me the code.
   Email: korapatimanoj@gmail.com
   Thanks in advance...

Regards,
Manu

What is in test 9?

RT~~

A simple problem , think carefully~

test 10?
And how can i watch the wrong answered tests?

Edited by author 13.09.2011 00:50

17
WEARECHAMPIONSYES
WEHAVECHAMPIONSNO

What should I write in this case ???
Sory my english


Edited by author 10.04.2011 23:15

Hello!
This is listing of my class

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;


public class q1496 {

    /**
     * @param args
     * @throws IOException
     * @throws NumberFormatException
     */
    public static void main(String[] args) throws NumberFormatException, IOException
    {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        int amount = Integer.parseInt(in.readLine());
        String[] source= new String[amount];
        String[] uniT= new String[amount];
        int j=0;
        int count=0;
        for (int i=0;i<amount;i++)
           {
            source[i]=in.readLine();
           }
        Arrays.sort(source);
        uniT[0]=source[0];

        for (int i=0;i<amount;i++)
           {
            if (source[i].equals(uniT[j])!=true)
               {
                j++;
                uniT[j]=source[i];
               }
           }
        for (int i=0;i<=j;i++)
            {
            count=0;
             for (int k=0;k<amount;k++)
                 {
                 if (source[k].equals(uniT[i]))
                     {
                     count++;
                     }
                   }
             if (count>1){System.out.println(uniT[i]);}
            }
    }

}

Hi there!
Look at my approach for solving this problem.
I`m using 1-dimensional representation of digital matrix. In example matrix M below:
       A11, A12, A13, A14
 (M) = A21, A22, A23, A24
       A31, A32, A33, A34
       A41, A42, A43, A44
transforms to the vector V:
 V = (A11, A12, A13, A14, A21, A22, A23, A24, A31, A32, A33, A34, A41, A42, A43, A44)
(array of integers in my C++ implementation).
Than point out the order in which elements of matrix M should appears in result:
(1)  (3)  (6)  (10) (2)  (5)  (9)  (13) (4)  (8)  (12) (15) (7)  (11) (14) (16)
A11, A12, A13, A14, A21, A22, A23, A24, A31, A32, A33, A34, A41, A42, A43, A44, where
values in brackets are elements serial numbers in result.

Onwards take a pencil and draw a curves between elements in desired order:
(1)  (3)  (6)  (10) (2)  (5)  (9)  (13) (4)  (8)  (12) (15) (7) (11) (14) (16)
A11, A12, A13, A14, A21, A22, A23, A24, A31, A32, A33, A34, A41, A42, A43, A44
 |___↑_|__↑_________↑ |  ↑ |            ↑ |
     |_|__|___________|  | |            | |
       |__|______________|_|____________| |
          |              |_|______________|
          |________________|                    ... and so on

In example above 6 elements ordered by curves connections. It should be observed that
it`s possible to fix sequence of actions using numeric relative offsets. See below:
Step  1) +4 (+4 is offset (in elements) of first element A11);
Step  2) -3 (-3 is offset of element, obtained in step 1);
Step  3) +7 (+7 is offset of element, obtained in previous step and so on);
Step  4) -3
Step  5) -3
Step  6) +10
Step  7) -3
Step  8) -3
Step  9) -3
Step 10) +10
Step 11) -3
Step 12) -3
Step 13) +7
Step 14) -3
Step 15) +4

Halves of steps sequence are simmetrical. Sequence contains two operation type: forward offset and backward offset (all offset are relative). Value of backward always equals to (side of matrix - 1). Value of forward offset different on difference steps. On first and last steps it`s equal to side size. On other steps it`s increased or decreased (above or below of middle steps) on value (side of matrix - 1).
Forward steps amount is always equals to 2 x (side-1). Amount of backwards steps are 1 on first forward step, 2 on second and so on toward the middle of iterations. After middle backwards steps amount decreased by 1 on each iterations except last. On last iterations it`s not backwards steps.

This algorithm has simple (not required complex data structures) implementation on various language.

Thank you for your attention and Good luck!

P.S. Sorry for my English =)

P.P.S. Graphical representation of algorithms steps is distorted because when message is sent any occurance of double, triple and so on spaces are replaced on one space character :(

Edited by author 13.09.2011 15:09

Hi,
as I understand, this is a problem on dynamic programming. I see only one solution - go through the whole tree and compute the values for each node from the leaves up to the root of the tree.
And on each level the full search should be done. For example, if we want to compute value for a node with p childs with already computed x_1, x_2, .., x_p values, then try 1/p, 2/p, ... p/p as a possible solution for this node.

please, give me any hint on this problem.

Edited by author 29.03.2011 20:27