The problem say that "It is guaranteed that during the experiment each time a coin was to be removed the stack was not empty."
But when I submit such a brute force code:
--------------------
        if (x > 0) a[t++] = x;
        else if (x == 0) {
            if (t + t < maxn) {
                copy(a, a + t, a + t);
                t += t;
            }
        }
        else printf("%d\n", a[--t]);
--------------------
It crashed at #42. Then, I modified the last "else branch" into:
--------------------
        else if (t) printf("%d\n", a[--t]);
--------------------
I got WA at #42.

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
char v[53];int ch[53];
char vi[53][5],*punt=NULL,vf[53],vc[53],vp[53],*a,vt[53],chr[3];
int maxl=0,maxt=0,k,b,max2,l=0;
int n,j=0,i,ind;

main()
{
    gets(v);
    n=strlen(v);
    if(n==1){cout<<v;}
    else{



        for(i=0;i<n-1;i++)
            {   // carga inicial
               vi[i][0]=v[i] ;
               vi[i][1]=v[i+1] ;
               ch[i]=0;


            }

        for(i=0;i<n-1;i++)
        {
            punt=strstr(v,vi[i]);


            if(punt!=NULL)
                {
                    ch[punt-v]++;


                }
        }
        for(i=0;i<n-1;i++)
        {

            if(ch[i]>maxt)
                    {
                        maxt=ch[i];
                        ind=i;

                    }



        }
        strcpy(vc,vi[ind]);
        strcpy(vp,v);

        punt=strstr(v,vi[ind]);
        i=0;
 strcpy(vf,vi[ind]);
        if (maxt!=0)
        {  while(1)

        {




            b=1;

            punt=strstr(v,vf);
            chr[0]=punt[2+l];

            strcat(vf,chr);



           k=2+l;
           k++;
           l++;
           if(l==strlen(punt))break;
           max2=0;
           strcpy(vp,v);
           while(1)
               {
                    punt=strstr(vp,vf);



                    if(punt!=NULL)
                        {

                            j=strlen(punt);

                            for(i=0;i<j-1;i++)
                            {
                                vp[i]=punt[i+1];
                            }
                            vp[i]='\0';



                            max2++;
                            if (max2>=maxt)
                            {
                                maxt=max2;
                                strcpy(vc,vf);
                                b=0;

                            }




                        }
                        if(punt==NULL)break;


                }
                if(b==3) break;


        }





        }
cout<<vc;

    }}

New tests were added to the problem. First tests were replaced with examples from the problem statements. 106 authors have lost AC after the rejudge.

I have used manacher'a algorithm   and tried all the test cases discussed in these discussions and getting correct solution....
I have used manacher for even and odd palindromes separately and found the longest and if longest comes again... as in the same length..take the smaller start..
but now,... I got WA #7 ...I initially faced some problem in compilation too..please suggest some test case

P.S. : m doing while(getline(cin,string))  and also tried while(cin>>string)
is something to be done here...please suggest

What in test #8?

Edited by author 25.07.2013 00:58

my method is DP:
//invite
int sum = 0;
for (Integer c : guest.children) {
    sum += DP[c.intValue()][0];
}
DP[index][1] = sum + RATE[index];

// do not invite
sum = 0;
for (Integer c : guest.children) {
    int max = Math.max(DP[c.intValue()][0], DP[c.intValue()][1]);
    sum += max;
}
DP[index][0] = sum;



at first, I got WA on 13#,
DP the tree should from leaf to root,
but at first I sort the guests by its children count,
the right way is sort the guests by depth

Edited by author 25.07.2013 09:50

#include <iostream>
#include <math.h>

using namespace std;
void main(void)
{
    int a,b,c;
    cin>>a>>b;
    c=0;
    c=a/2+a%2+b/2+b%2;

    if (((a==1)&&(b!=1))||((a!=1)&&(b==1)))
    {
            c=c-1;
    }
    if (c%2) cout <<"[:=[first]" ;
    else cout <<"[second]=:]" ;

}
whot is wrong?

Edited by author 25.07.2013 00:45

Please give test 2 for this problem.

I take the list (any City with 1 road) and delete this road, then going to connected city and mark all road from it as non-delete, then  i going to near city and repeat first step.
For example
5 4
1 2
2 3
3 4
3 5
Start with 1 City, delete (1 2) then go in 3 city, and remove (3 4) or (3 5)?

Why WA2 ?
Sorry fo my English.

    while(scanf("%d%d",&able,&mytime)!=EOF)
    {
        sum=0;
        while(mytime--)
        {
            scanf("%d",&temp);
            sum+=(temp-able);
        }
        if(sum<=0)
            printf("%d\n",0);
        else
            printf("%d\n",sum);
    }

Please give me test. Thank you!

it seems like test#4 includes big latin letters, but it's impossible according problem statement "все слова записаны в нижнем регистре, слово может содержать только латинские буквы от 'a' до 'z' (то есть «one-seventh» — это два слова — «one» и «seventh»)."
admin, please check

>>500
732986521245023


:)

#include<stdio.h>
#include<stdlib.h>
int compare(const void *a, const void *b)
{
    if(*(*(int**)a+1)<*(*(int**)b+1))
        return 1;
    else
        return -1;
}
int main()
{
    int n,i,**arr;
    scanf("%d",&n);
    arr=(int**)calloc(n,sizeof(int*));
    for(i=0;i<n;i++)
    {
        arr[i]=(int*)calloc(2,sizeof(int));
        scanf("%d %d",&arr[i][0],&arr[i][1]);
    }
    qsort(arr,n,4,compare);
    for(i=0;i<n;i++)
    {
        printf("%d %d\n",arr[i][0],arr[i][1]);
        free(arr[i]);
    }
    free(arr);
    return 0;
}

A bug or feature the situation that
last <ENTER> does not count as an error??

For example on test

"
12
un-usual
for-ced
re-al
sum-mer
N-ED
home-work
sec-ret
ho-lidays
Wi-zard
th-Er
L-so
Holi-days
Harry Potter was a highly unusual boy in many ways.
For one thing, he hated the summer holidays more than any
other time of year. For another, he really
 wanted to do his homework but was forced to do it in
secret, in the dead of night. And he also happened
to be a wizard.
"

my program gives away
"
Harry Potter was a highly unusual boy in
 many ways.
For one thing, he hated the summer holi-
days more than any
other time of year. For another, he
really
 wanted to do his homework but was for-
ced to do it in
secret, in the dead of night. And he al-
so happened
to be a wizard.<ENTER>
"
(All the attention on line carry after "wizard.")

I've done it to pass WA#9, cause understood
that
output:
"
"
(nothing)

on input like
"
<ENTER>
"
not good...

Edited by author 23.07.2013 03:14

Edited by author 23.07.2013 03:14

import java.util.*;
import java.io.*;

public class p1656
{
    public static int reverse(int k, int N)
    {
        return N - k - 1;
    }

    public static point fun(int i, int j, int N, int function)
    {
        point p = new point();
        if (function == 1)
        {
            if (i != j)
            {
                p.i = reverse(j, N);
                p.j = reverse(i, N);
            } else
            {
                p.i = reverse(i, N);
                p.j = j;
            }
        } else if (function == 2)
        {
            if (i != j)
            {
                p.i = j;
                p.j = i;
            } else
            {
                p.i = i;
                p.j = reverse(j, N);
            }
        } else if (function == 3)
        {
            if (i != j)
            {
                p.i = reverse(i, N);
                p.j = reverse(j, N);
            } else
            {
                p.i = reverse(i, N);
                p.j = reverse(j, N);
            }
        }
        return p;
    }
    public static void main(String[] args)
    {
        Scanner in = new Scanner(System.in);

        int N = in.nextInt();

        int arr[] = new int[N * N];
        //{198, 186, 181, 190, 194, 194, 174, 188, 171, 198, 182, 171, 175, 184, 177, 200, 189, 185, 177, 171, 172, 180, 186, 188, 197};
        int[] out = new int[N * N];

        for (int i = 0; i < N * N; i++)
        {
            arr[i] = in.nextInt();
        }

        Arrays.sort(arr);

        /*for (int i = 0; i < N * N; i++)
            System.out.print(arr[i] + " ");
        System.out.println()*/;

        int po = 0;

        for (int j = 0; j < N / 2 ; j++)
        {
            for (int i = j; i < N / 2 + 1; i++)
            {
                // System.out.println("i = " + i + " j = " + j + " po = " + po);

                out[N * i + j] = arr[po++];
                point p = fun(i, j, N, 1);

                //System.out.println("i = " + p.i + " j = " + p.j + " po = " + po);

                out[N * p.i + p.j] = arr[po++];
                p = fun(i, j, N, 2);

                //System.out.println("i = " + p.i + " j = " + p.j + " po = " + po);

                out[N * p.i + p.j] = arr[po++];
                p = fun(i, j, N, 3);

                 //System.out.println("i = " + p.i + " j = " + p.j + " po = " + po);

                out[N * p.i + p.j] = arr[po++];

                if (i != j && i != N / 2 )
                {
                    int savei = i;
                    i = reverse(i, N);


                    // System.out.println("i = " + i + " j = " + j + " po = " + po);

                    out[N * i + j] = arr[po++];
                    p = fun(i, j, N, 1);

                     //System.out.println("i = " + p.i + " j = " + p.j + " po = " + po);

                    out[N * p.i + p.j] = arr[po++];
                    p = fun(i, j, N, 2);

                    // System.out.println("i = " + p.i + " j = " + p.j + " po = " + po);

                    out[N * p.i + p.j] = arr[po++];
                    p = fun(i, j, N, 3);

                    // System.out.println("i = " + p.i + " j = " + p.j + " po = " + po);

                    out[N * p.i + p.j] = arr[po++];

                    i = savei;
                }

            }
        }

        out[N / 2  * N + N / 2] = arr[po++];

        for (int i = 0; i < N; i++)
        {
            for (int j = 0; j < N; j++)
                System.out.format("%4d", out[i * N + j]);
            System.out.println();
        }
    }
}

class point
{
    int i, j;
}

If the test is:

4 6
1 2
1 3
1 4
2 3
2 4
3 4

then all created routes are as follows:

1 - 2 - 3 - 1
1 - 2 - 4 - 1
1 - 3 - 4 - 1
2 - 3 - 4 - 2
1 - 2 - 3 - 4 - 1
1 - 2 - 4 - 3 - 1
1 - 3 - 2 - 4 - 1

Am I right?

#include <iostream>


using namespace std;
int main()
{
    int n, i, j;
    cin>>n;
    int a[n], s=0, s1=0, s2=0, s3=0, s4=0, s5=0, s6=0, s7=0, s8=0, s9=0, s10=0;
    for(i=1; i<=n; i++)
    cin>>a[i];
    for(i=1; i<=n; i++)
    {

             if(a[i]==1) s1++;
             if(a[i]==2) s2++;
             if(a[i]==3) s3++;
             if(a[i]==4) s4++;
             if(a[i]==5) s5++;
             if(a[i]==6) s6++;
             if(a[i]==7) s7++;
             if(a[i]==8) s8++;
             if(a[i]==9) s9++;
             if(a[i]==10) s10++;
             //if(a[i]==0) s++;


    }
    if(s1==0) cout<<"";
    else cout<<s1<<" "<<"1 ";
    if(s2==0) cout<<"";
    else cout<<s2<<" "<<"2 ";
    if(s3==0) cout<<"";
    else cout<<s3<<" "<<"3 ";
    if(s4==0) cout<<"";
    else cout<<s4<<" "<<"4 ";
    if(s5==0) cout<<"";
    else cout<<s5<<" "<<"5 ";
    if(s6==0) cout<<"";
    else cout<<s6<<" "<<"6 ";
    if(s7==0) cout<<"";
    else cout<<s7<<" "<<"7 ";
    if(s8==0) cout<<"";
    else cout<<s8<<" "<<"8 ";
    if(s9==0) cout<<"";
    else cout<<s9<<" "<<"9 ";
    if(s10==0) cout<<"";
    else cout<<s10<<" "<<"10 ";
    cout<<endl;
"<<s8<<" 8 "<<s9<<" 9 "<<s10<<" 10 ";





}