Here is my code

#include <stdio.h>
#define MaxN 100000+252
#define MaxPop MaxN/2+1


struct pointer {
    int s;
    char o;
};

struct pointer GetPointer(int i)
{
    struct pointer p;
    p.s=i/250;
    p.o=i % 250;
    return p;

}

struct Stack {
    int V;
    struct pointer pred;
};
int GetAdress(struct pointer p) {
    return p.s*250+p.o;
}

int A[1001];
int MP;
struct Stack B[MaxN];


int Pop(int S)
{
    int E;
    E=B[A[S]].V;
    A[S]=GetAdress(B[A[S]].pred);
    return E;
}
int Push(int S, int E)
{
    int i=A[S];
    MP++;
    B[MP].V=E;
    B[MP].pred=GetPointer(i);
    A[S]=MP;
    return 0;
}
int main() {

    int N,i,j,E,k;
    int S;
    char command[10];
    scanf("%d",&N);


    MP=0;
    B[MP].V=0;
    B[MP].pred=GetPointer(0);
    for (i=1;i<1000;i++) A[i]=0;
    k=0;
    for (i=1;i<=N;i++) {

        scanf("%s",command);
        if (command[1] == 'U'){
            scanf("%d %d",&S,&E);
            Push(S,E);
        } else {
            scanf("%d",&S);
            k++;
            printf("%d\n",Pop(S));
        }
    }




    return 0;
}

Edited by author 10.02.2013 18:10

What is test 6? I am stuck at it for a very long time!

What is the true way to change k-(needed numbers)
http://pastebin.com/f27Pt0HC - code
#include <iostream>;
#include <map>;
#include <string>;
using namespace std;
// dfs( копия матрицы, кол-во вершин, указатель на юзет, в который будут добавлятся числа Исенбаева, проверяемое ребро, число Исенбаева в данный момент)
int dfs(int **gr, int n, int *&q, int v, int k)
{
    for(int i=0;i<n;i++)
        if ( (gr[v][i]==-1)&&(q[i]==1000) )//если из v в i есть путь и если i не обойдена
        {
            q[i]=min(k,q[i]);//чтобы избавится от 1000
            dfs(gr,n,q,i,k+1);
        }
    return 1;
}


int main()
{

    int x,i,j,m,n;

    map<string,int> a;
    n=0;

    cin>>m;
    int **gr=new int*[3*m]; //матрица смежности
        for(int i=0;i<3*m;i++)
            gr[i]=new int[3*m];
    string s1,s2,s3;
    for(int i=0; i<m; i++)//ввод фамилий по 3 в строке
    {
        cin>>s1>>s2>>s3;
        if (a[s1]==0)
            a[s1]=++n;
        if (a[s2]==0)
            a[s2]=++n;
        if (a[s3]==0)
            a[s3]=++n;
        gr[ a[s1] ] [ a[s2] ] =-1;
        gr[ a[s2] ] [ a[s1] ] =-1;
        gr[ a[s3] ] [ a[s1] ] =-1;
        gr[ a[s1] ] [ a[s3] ] =-1;
        gr[ a[s3] ] [ a[s2] ] =-1;
        gr[ a[s2] ] [ a[s3] ] =-1;
    };
    int k=1;
    x=a["Isenbaev"];
    int *used=new int[n];
    for(int i=0;i<n;used[i++]=1000);
    int *&q=used;
    cout<<dfs(gr,n,used,x,1);
    i=0;
    used[x]=0;
    for(map<string,int>::const_iterator x=a.begin();x!=a.end();++x,i++)
        if(used[i]<1000)
            cout<<x->first<<' '<<used[i]<<endl;
        else
            cout<<x->first<<" undefined"<<endl;

    delete gr,used;
    return 0;
}

Could anyone give me the test#6?
godsuriyel@gmail.com
thanks!

/*
 * You should only define the SURC_TEST macros
 * in the compile directives!
 */
#include <stdio.h>

void main(void)
{
    int nT, a, b, nResult;
#ifndef SURC_TEST
    scanf ("%d", &nT);
#else
    nT = 27;
#endif
    for (int iT = 0; iT < nT; ++iT)
    {
#ifndef SURC_TEST
        scanf ("%d %d", &a, &b);
#else
        switch (iT)
        {
        case 0:
            a = 30;
            b = 89;
            break;
        case 1:
            a = 2;
            b = 16;
            break;
        case 2:
            a = 3;
            b = 243;
            break;
        case 3:
            a = 1;
            b = 1;
            break;
        case 4:
            a = 2;
            b = 2;
            break;
        case 5:
            a = 2;
            b = 72;
            break;
        case 6:
            a = 1;
            b = 536870912;
            break;
        case 7:
            a = 1;
            b = 536870912;
            break;
        case 8:
            a = 1;
            b = 214358881;
            break;
        case 9:
            a = 1;
            b = 147008443;
            break;
        case 10:
            a = 1;
            b = 352275361;
            break;
        case 11:
            a = 1;
            b = 28934443;
            break;
        case 12:
            a = 1;
            b = 214921799;
            break;
        case 13:
            a = 1;
            b = 1042441;
            break;
        case 14:
            a = 1;
            b = 2627641;
            break;
        case 15:
            a = 1;
            b = 6086089;
            break;
        case 16:
            a = 1;
            b = 12823561;
            break;
        case 17:
            a = 1;
            b = 24730729;
            break;
        case 18:
            a = 1;
            b = 44930209;
            break;
        case 19:
            a = 1;
            b = 77810041;
            break;
        case 20:
            a = 1;
            b = 129254161;
            break;
        case 21:
            a = 1;
            b = 207561649;
            break;
        case 22:
            a = 1;
            b = 319730161;
            break;
        case 23:
            a = 1;
            b = 480091921;
            break;
        case 24:
            a = 1;
            b = 707081281;
            break;
        case 25:
            a = 1;
            b = 31973;
            break;
        case 26:
            a = 1;
            b = 499999999;
            break;
        default:
            a = 1;
            b = 1;
            break;
        }
#endif
        nResult = 0;
        /********** ALGORITM START ********
         * Insert your algorithm here.
         * input: a, b
         * result write to: nResult
         */

        // ********* ALGIROTM END *********
#ifndef SURC_TEST
        printf ("%d\n", nResult);
#else
        int nRightResult;
        switch (iT)
        {
        case 0:
            nRightResult = 0;
            break;
        case 1:
            nRightResult = 4;
            break;
        case 2:
            nRightResult = 5;
            break;
        case 3:
        case 4:
            nRightResult = 1;
            break;
        case 5:
            nRightResult = 5;
            break;
        case 6:
        case 7:
            nRightResult = 30;
            break;
        case 8:
            nRightResult = 9;
            break;
        case 9:
            nRightResult = 6;
            break;
        case 10:
            nRightResult = 5;
            break;
        case 11:
        case 12:
            nRightResult = 4;
            break;
        case 13:
        case 14:
        case 15:
        case 16:
        case 17:
        case 18:
        case 19:
        case 20:
        case 21:
        case 22:
        case 23:
        case 24:
            nRightResult = 3;
            break;
        case 25:
            nRightResult = 2;
            break;
        case 26:
            nRightResult = 3;
            break;
        default:
            nRightResult = 1;
            break;
        }
        if (nResult == nRightResult)
            printf ("%d\tT\n", iT);
        else
            printf ("%d\tF: %d (should be: %d)\n", iT, nResult, nRightResult);
#endif
    }
}

Edited by author 09.02.2013 22:15

Edited by author 09.02.2013 22:15

who know 13 test . Please help me

If you use dfs, try this test

5
1 3
01000
10111
01000
01000
01000

Answer is:

0
00000
00000
00000
00000
00000

Some tests

7 2
1 2 3 4 5 6 7
Answer
1 5
2 6
3 7
4


7 6
1 2 3 4 5 6 7
Answer
1 3 4 5 6 7
2

ALL tests from the discussion passed and still WA7. Any suggestions?

The key for solution was realizing that the ends of the sticks was on the same distance from origin. How to prove that?
I have a function F(angleA, angleB). I took derivatives dF/angleA and dF/angleB, they must be equal to zero. Then what?

var i,j,k,n:longint;
begin
 read(n);

  for i:=1 to 250 do
   if sqr(i)=n then begin write(1); exit; end;

  for i:=1 to 250 do
   for j:=1 to 250 do
    if sqr(i)+sqr(j)=n then begin write(2); exit; end;

  for i:=1 to 250 do
   for j:=1 to 250 do
    for k:=1 to 250 do
     if sqr(i)+sqr(j)+sqr(k)=n then begin write(3); exit; end;

write(4);
end.

# include <math.h>
# include <stdio.h>
# include <algorithm>

using namespace std;

int a,b,c,x,y;

int main()
{
    scanf("%d %d %d",&a,&b,&c);

    for(int i=1;i<10001;i++)
    {
        x=0;
        y=0;

        for(int j=1;j<=(int)sqrt((double)i)+1;j++)
        if(i%j==0)
        x++;

        if(i-c>0)
        {
            for(int j=1;j<=(int)sqrt((double)(i-c))+1;j++)
            if(i%j==0)
            y++;
        }

        if(b==x && a==y)
        {printf("%d",i);getchar();getchar();return 0;}
    }

    printf("0");system("Pause");
}

//SESC USU SUDO RM -RF

var a,b,c,d,i:integer;
begin
readln(a);
readln(b);
for i:=1 to 10000 do
  begin
    if i mod 2 = 0 then
    if b-c <> 0 then
    c:=c+1
      else d:=d+1
    else if a-c <> 0 then
    c:=c+1
      else d:=d+1;
  end;
case d of
  1..200000 : writeln ('yes');
  0 : writeln ('no');
  end;
end.

Edited by author 01.12.2012 17:39

Edited by author 01.12.2012 17:39

Edited by author 01.12.2012 17:39

var   a,b:integer;

begin
   if(a mod 2 =1) or (b mod 2=0)  then writeln('no')
   else writeln('yes')
end.

or wa1?


var c,d : integer;
    a,b : string[4];
begin
   readln(a);
   readln(b);
   val(a,c);
   c:=c mod 2;
   val(b,d);
   d:=d mod 2;
   writeln(c,d);
   if (c=1) or (d=0) then writeln('no')
   else writeln('yes');

end.

1
0

Must be "Death"

really curious about it, do you have any idea?
Am I right?
19
1
1 1
10
6 4
100
60 12
1000
840 32
10000
7560 64
100000
83160 128
1000000
720720 240
10000000
8648640 448
100000000
73513440 768
1000000000
735134400 1344
10000000000
6983776800 2304
100000000000
97772875200 4032
1000000000000
963761198400 6720
10000000000000
9316358251200 10752
100000000000000
97821761637600 17280
1000000000000000
866421317361600 26880
10000000000000000
8086598962041600 41472
100000000000000000
74801040398884800 64512
1000000000000000000
856811873911587456 491520

I can't understand what's wrong?
#include <iostream>
using namespace std;
int main()
{
    int n,s=1,max,j,i;
    cin>>n;
    int *a=new int[n];

    for(i=0; i<n;i++)
        cin>>a[i];
    s=1;max=0;
    for(i=0;i<n;i++)
    {
        j=i+1;
        while(a[i]<a[j])
            j++;
        if (j-i>=s)
        {
            if (max>0)
                s=j-i+1;
            else
                s=j-i;
            max=i;
        }
    };
    cout<<max+1;




    delete a;
    return 0;
}

Edited by author 07.02.2013 01:30

In paskal when n=10000 you can not write this expression n*(n+1)*(n+2)/2 in some variables it is very long and big number so in C++ it is very easy just write(here is my AC program):
#include <cstdlib>
#include <iostream>

using namespace std;

int main(int argc, char *argv[])
{
    long long n;
    cin>>n;
    cout<<n*(n+1)*(n+2)/2<<endl;
    system("PAUSE");
    return EXIT_SUCCESS;
}

Edited by author 27.04.2008 17:47