how can i see my current accepted submission rank????

#include<iostream>
using namespace std;
int n,i,k,t,j;
string a[10000],b[10000],c[10000],d[10000];
int main(){
        a[1]="Alice"; a[4]="Phil";  a[7]="Phoebus";
        a[2]="Ariel"; a[5]="Peter";  a[8]="Ralph";
        a[3]="Aurora"; a[6]="Olaf";     a[9]="Robin";

        b[1]="Bambi"; b[4]="Mulan";  b[7]="Silver";
        b[2]="Belle"; b[5]="Mowgli";  b[8]="Simba";
        b[3]="Bolt";  b[6]="Mickey";  b[9]="Stitch";

        c[1]="Dumbo"; c[4]="Kuzko"; c[7]="Tarzan";
        c[2]="Genie"; c[5]="Kida";  c[8]="Tiana";
        c[3]="Jiminy"; c[6]="Kenai"; c[9]="Winnie";
        cin>>n;
            for(i=1; i<=n; i++) cin>>d[i];
                for(i=1; i<=n; i++) {
                    for(j=1; j<=9; j++)  if(d[i]==a[j]){
                    if(t==0) k+=0;
                    if(t==1) k+=1;
                    if(t==2) k+=1;
                    if(t==3) k+=2;
                    t=1;
                    break;
                    }
                    for(j=1; j<=9; j++)  if(d[i]==b[j]){
                    if(t==0) k+=0;
                    if(t==1) k+=1;
                    if(t==2) k+=1;
                    if(t==3) k+=1;
                    t=2;
                    break;
                    }
                    for(j=1; j<=9; j++) if(d[i]==c[j]) {
                    if(t==0) k+=0;
                    if(t==1) k+=2;
                    if(t==2) k+=1;
                    if(t==3) k+=1;
                     t=3;
                     break;
                     }
}
            cout<<k;
}

what's wrong with this code:
#include <iostream>
#include <vector>
#include <list>
using namespace std;

int ar[102][102],viz[102];
int n;
list <int> a;
int dfs(int s)
{
    viz[s] = 1;
    for (int i=1;i<=n;i++)
    {

        if (ar[i][s] && !viz[i]) {dfs(i);  }//a.push_back(i);}
    }
    return 0;
}

int main()
{
    cin >> n;
    int z;
    for (int i=1;i<=n;)
    {
        cin >> z;
        if (z == 0) i++;
        else ar[i][z] = 1;
    }
    dfs(1);
    for (int i=1;i<=n;i++)
        if(!viz[i]) a.push_back(i);
    cout << a.size() << endl;
    for (int n : a) {
        cout << n << " ";
    }
    return 0;
}


maybe i didn't understand problem properly,some help?

Here is my solution and used test cases. I don't understand why I
always get WA. 8-(

Test 1.
5 4 2 0 1 2: NO

Test 2.5 4 2 0 1 5: YES

Test 3.
5 4 2 0 3 5: YES

Test 4.
5 4 2 0 3 4: YES

Test 5.
5 5 2 0 3 4 5: YES

Test 6.
5 5 2 0 2 4 5: YES

Test 7.
5 6 2 1 2 4 5 0: NO

Test 8.
5 4 1 2 2 3: YES

Test 9.
5 5 1 2 3 4 3: YES

Test 10.
5 5 1 1 0 3 4: NO

Test 11.
5 3 0 0 1: NO

Test 12.
5 3 0 1 1: NO

Test 13.
5 3 1 2 2: YES

Test 14.
5 3 1 1 2: YES

Test 15.
5 3 4 4 5: NO

Test 16.
5 3 3 4 4: YES

Test 17.
5 3 4 5 5: NO

Test 18.
5 3 3 4 5: YES

Test 19.
2 2 1 1: YES

Test 20.
2 2 2 2: NO

Source code:

[code deleted]

Edited by moderator 13.02.2007 20:57

there are n * m possible sums. to find a solution, we can try to create such vectors such that all n * m sums are different.
Suppose the 1st vector is 1 2 3 ... n

Then the first element of next vector is taken to be n + 1(we require all unique).
Now we need to find the next element while making all n*p sums different, where p is the size of the 2nd vector.

n+1
1 2 3 ... n

initially all sums(n+2 to 2n+1) are distinct. Now we need to add new element such that
smallest possible new sum is greater than the previous sum
i.e. if new element is p
p+1>2n+1
or p >= 2n + 1
in general p+1>prev+n
p>=prev+n

what is wrong here?

for (int i =0; i< 15;++i)
{
int p = primes[i];
if ( c % p == 0)
{
   unsigned long long M = c / p;
   unsigned long long M_inv = inverse[i][ M % p ] ;
   unsigned long long h = M_inv * m[ i ];
   // r = (r + h * M ) % c

   unsigned long long high_M = (M >> 32 ) & 0xFFFFFFFFULL;
   unsigned long long low_M = ( M & 0xFFFFFFFFULL);

   // h * M= h * (high_M * 2^32 + low_M)
   unsigned long long high_h = high_M * h;
   unsigned long long e = (high_h >> 32) & 0xFFFFFFFFULL;
   unsigned long long f = (high_h & 0xFFFFFFFFULL);

   // high_h = e*2^32 + f

// h*M = h*(high_M*2^32 + low_M) = high_h * 2^32 + low_M * h =
// = (e*2^32 + f)*2^32 + low_M * h = e*2^64 + f*2^32 + low_M * h

    e = (e << 48) % c;
    e = (e << 3 ) % c;
    e = (e << 3 ) % c;
    e = (e << 3 ) % c;
    e = (e << 3 ) % c;
    e = (e << 3 ) % c;
    e = (e << 1 ) % c;

   e = (e + (f << 32) ) % c;
   e = (e + low_M * h ) % c;

   r = (r + e ) % c;
 }
}

You have to divide the participants into equal teams (rounded)
For example, for "15 10" test - (1, 1, 1, 1, 1, 2, 2, 2, 2, 2)
Good luck :)

int n = in.nextInt();
        int l = 0;
        if (n > 24) {
            out.print("Glupenky Pierre");
        } else {
            for (int i = 1; i <= n; i++) {
                if (i / 5 == 0)
                    l = 0;
                if (i / 5 == 1)
                    l = 1;
                if (i / 5 == 2)
                    l = 8;
                if (i / 5 == 3)
                    l = 6;
                if (i / 5 == 4)
                    l = 9;

                if (i % 5 == 0)
                    l = l * 10;
                if (i % 5 == 1)
                    l = l * 10 + 1;
                if (i % 5 == 2)
                    l = l * 10 + 8;
                if (i % 5 == 3)
                    l = l * 10 + 6;
                if (i % 5 == 4)
                    l = l * 10 + 9;
                if (l/10==0)
                    out.print("0");
                out.print(l);
                out.print(" ");
            }
        }

Edited by author 07.11.2016 22:26

Edited by author 07.11.2016 22:27

There is my code:

Program t1134;

Const MaxN=1000;

Var Card      : array[1..MaxN]of boolean;
    Numb      : array[1..MaxN]of integer;
    Use       : array[1..MaxN]of boolean;
    M,N,i,j   : integer;
    ex        : boolean;

begin
Read(N,M);
if (M>N)or(m=0) then begin
 writeln('NO');
 halt(0);
end;
for i:=1 to N do Card[i]:=false;
for i:=1 to M do Use[i]:=false;
for i:=1 to M do begin
 read(Numb[i]);
 if (Numb[i]<0)or(Numb[i]>n) then begin
   writeln('NO');
   halt(0);
 end;
end;
for i:=1 to M do
 if Numb[i]=0 then begin
  Use[i]:=true;
  if Card[1] then begin
    writeln('NO');
    halt(0);
  end;
  Card[1]:=true;
 end else
 if Numb[i]=n then begin
  Use[i]:=true;
  if Card[n] then begin
    writeln('NO');
    halt(0);
  end;
  Card[n]:=true;
 end;
for i:=1 to m-1 do if not(use[i]) then
 for j:=i+1 to m do if not(use[j]) then
  if numb[i]=numb[j] then begin
   if (card[numb[i]])or(card[numb[j]-1]) then begin
    writeln('NO');
    halt(0);
   end;
   card[numb[i]-1]:=true;
   card[numb[i]]:=true;
   use[i]:=true;
   use[j]:=true;
  end;
repeat
ex:=true;
for i:=1 to m do if not(use[i]) then
 if (card[numb[i]])or(card[numb[i]-1]) then begin
  if (card[numb[i]])and(card[numb[i]-1]) then begin
    writeln('NO');
    halt(0);
  end;
  if card[numb[i]] then begin
   card[numb[i]-1]:=true;
   use[i]:=true;
  end else begin
   card[numb[i]-1]:=true;
   use[i]:=true;
  end;
 end;
until ex;
Writeln('YES');
end.

does anyone knows test 6?

http://pastebin.com/af0DFBg0
Look at my code please.
If someone could say me what wrong, it would be great.
Thank you!

У задачи отклеилось описание на русском? Хоть и шарю по английскому, всё равно неприятно тратить несколько минут на перевод...

Edited by author 20.02.2016 16:49

i see N%3 solution everywhere but does anyone have any explanation as to why the guy who has mod-3 no of stones should lose. Some logical or mathematical reason would be much appreciated.

why should i declare 128*1024 size of memory for 256KB described in the problem?
my code goes here:
#include<iostream>
#include <cstdio>
#include <cmath>

double buf[128 * 1024];
int main()
{
    int i = 0;
    unsigned long long n;
    while (scanf("%llu", &n) != EOF) {
        buf[i ++] = (double)sqrt(n * 1.0);
    }
    for (; --i >= 0; ) {
        printf("%.4lf\n", buf[i]);
    }
    return 0;
}

//  why 128*1024  ?

Hello,
I've not found a topic about this test. Does anyone know what is it?
I'm just searching from the half of the string to find where can be the center of a palindom.
Thank you

The follwing is my pgm. Could some give a test case not satisfied here?
var
s1,i:integer;
f:boolean;
a2,b2,a1,b1,a,b:real;
begin

 f:=false;
 read(a);
 read(b);
 a1:=a/100;
 b1:=b/100;
 if b<>0 then
 i:=trunc(1/b1);
 a2:=a1*(i);
 b2:=b1*(i);
 while not f do
    begin
   if (((trunc(b2)<>trunc(a2)))and (frac(b2)<>0)   then f:=true;
   a2:=a2+a1;
   b2:=b2+b1;
   i:=i+1;
 end;
 i:=i-1;
 writeln(i);

end.

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>

const long long mod=1000000007;
const int nmax=200;

int n,m,t,p;
long long v[nmax],fact[nmax],step[nmax];
bool used[nmax];
bool a[nmax][nmax];
long long ans;

void DFS(int v)
{
       for(int j=1;j<=n;j++)
       {
              if(a[v][j] && !used[j])
              {
                     t++;
                     used[j]=true;
                     DFS(j);
              }
       }
}

bool FindCycle(int v,int pr)
{
     bool f=false;
     for(int j=1;j<=n;j++)
     {
            if(a[v][j] && j!=pr)
            {
                   if(used[j]) return true;
                   else
                   {
                     used[j]=true;
                     t++;
                     f=f | FindCycle(j,v);
                   }
            }
     }
     return f;
}

void main()
{
       scanf("%d%d\n",&n,&m);

        if(n<=2)
        {
               printf("1");
               return;
        }

       fact[0]=1;
       for(int i=1;i<=n;i++)
              fact[i]=fact[i-1]*((long long)i) % mod;

       for(int i=1;i<=n;i++) step[i]=0;

       for(int i=1;i<=n;i++)
       {
              for(int j=1;j<=n;j++)
                     a[i][j]=false;
       }

       for(int i=1;i<=m;i++)
       {
          scanf("%d\n",&t);
          if(!a[i][t])
          {
              a[i][t]=a[t][i]=true;
              step[i]++;
              step[t]++;

          }
       }

       p=1;
       while(p<=n && step[p]<=2) p++;
       if(p<=n)
       {
              printf("0");
              return;
       }

       for(int i=1;i<=n;i++) used[i]=false;

       for(int i=1;i<=n;i++)
       {
              if(!used[i])
              {
                   used[i]=true;
                   t=1;
                   if(FindCycle(i,-1))
                   {
                          if(t<n)
                          {
                            printf("0");
                            return;
                          }
                          if(t==n)
                          {
                            printf("2");
                            return;
                          }
                   }

              }
       }

       for(int i=1;i<=n;i++) used[i]=false;
       p=0;
       ans=1;
       for(int i=1;i<=n;i++)
       {
              if(!used[i])
              {
                     p++;
                     t=1;
                     used[i]=true;
                     DFS(i);
                     if(t>1) t=2;
                     ans*=((long long)t) % mod;
              }
       }
       ans=ans*fact[p-1] % mod;
       while(ans<0) ans+=mod;
       printf("%lld",ans);
}

a a b a a
answer: 1

If you have WA4 java. Put 10^7  instead of 10^6