WA #2

a = input().split()
s = input().split()
n, k = int(a[0]), int(a[1])
out = ["" for x in range(n//k+1)]
for i in range(k):
    if i != k - 1:
        for j in range(n//k+1):
            if len(s[0]) == 1:
                out[j] += "   " + s[0]
                s.pop(0)
            elif len(s[0]) == 2:
                out[j] += "  " + s[0]
                s.pop(0)
            else:
                out[j] += " " + s[0]
                s.pop(0)
    else:
        for j in range(n%k):
            if len(s[0]) == 1:
                out[j] += "   " + s[0]
                s.pop(0)
            elif len(s[0]) == 2:
                out[j] += "  " + s[0]
                s.pop(0)
            else:
                out[j] += " " + s[0]
                s.pop(0)
for i in out:
    print(i)

Edited by author 09.03.2018 19:59

import java.util.Scanner;

public class _1506 {
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int n, k, j, i;

        n = s.nextInt();
        k = s.nextInt();

        k = (n + k - 1) / k;
        int[] a = new int[n];
        for (i = 0; i < n; i++) {
            a[i] = s.nextInt();
        }

        for (j = 0; j < k; System.out.println(""), j++)
            for (i = j; i < n; i += k) {
                System.out.print(" " + a[i]);
            }
    }
}

var
     mas: array [1..100] of integer;
     mas2: array [1..100, 1..100] of integer;
     n, k, str, f, i, j, kol: integer;

begin
     readln(n, k);
        {for i:=1 to  n do read mas[i];}
        {Считаем кол-во строк}
     if n mod k <> 0 then str := (n div k) + 1 else str := n div k;
        {Считаем "полную" матрицу}
     kol := k * str;
     for i := 1 to n do read(mas[i]);
        {Заполняем "-1" будущие пустые места}
     for i := n + 1 to kol do mas[i] := -1;
     f := 1;
     for i := 1 to k do
          for j := 1 to str do
          begin
               mas2[j, i] := mas[f];
               inc(f);
          end;
     for i := 1 to str do
     begin
          for j := 1 to k do
               if mas2[i, j] <> -1 then write(mas2[i, j]:4) else continue;
          writeln;
     end;
end.

input :
15 4
1 2 30 994 50 600 700 999 900 990 991 992 993 40 800
output :
   1  50 900 993
   2 600 990  40
  30 700 991 800
 994 999 992

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

C++, 0.015 sec, 196 Kb

#include <iostream>
using namespace std;

int main()
{
    int n,k;
    cin>>n>>k;
    int m=n/k + ((n%k==0) ? 0:1);

    int * A=new int [m*k];
    for (int i=0;i<n;i++)
    {
        cin>>A[i];
    }
    for (int i=n;i<m*k;i++)
        A[i]=-1;

    for (int i=0;i<m;i++)
    {
        for (int j=0;j<k;j++)
        {
            int h=j*m+i;
            if (A[h]<0)
                cout<<"    ";
            if (A[h]>-1 && A[h]<10)
                cout<<"   "<<A[h];
            if (A[h]>9 && A[h]<100)
                cout<<"  "<<A[h];
            if (A[h]>99 && A[h]<1000)
                cout<<" "<<A[h];
        }
        cout<<endl;
    }

    return 0;
}

Waiting of comments

100 1
1 2 3 4 .. 99 100

ans:
1 2 3 4 .. 99 100


Edited by author 09.07.2011 17:54

What's incorrect?????

int main()
{
    int n, k;
    scanf("%d %d", &n, &k);

    vector<int> mas(100);

    mas.assign(100, -1);
    for (int i=0; i<n; i++)
        scanf("%d", &mas[i]);
    int s=n/k;
    if(n%k!=0) s++;

    for (int i=0; i<s; i++)
    {
        for(int j=0; j<k; j++)
            if (mas[s*j+i]!=-1)
            printf("%4d", mas[s*j+i]);
        printf("\n");
    }

  return 0;
 }

Edited by author 26.04.2011 20:30

I get WA#1
here is my code

#include <cstdlib>
#include <iostream>

using namespace std;
int i, j, n, k, say, a[101];

int main(int argc, char *argv[])
{
            cin >> n >> k;

                for( i = 0; i < n ; i++ )
                    cin >> a[ i ];
                   if( n%k != 0)
                    say = n/k + 1;
        else
            say = n/k;

                for( j = 0; j < say; j++ ){
                        cout << a[ j ];
                        i = j + say;
                        while( i < n){

                            cout <<" "<< a[ i ];
                            i += say;
                        }
                        if( j < say-1 )
                        cout << endl;
                        }



    return 0;
}

4 3
1 2 30 40

#include <stdio.h>

int x[101];
void main()
  {
    int n=0,m=0,i=0,j=0,k=0;
    scanf("%d", &n);
    scanf("%d", &m);
    i=0;
    for (i=0;i<n;i++) scanf("%d", &x[i]);
    i=0;
    m=2;
    while ((i<n)&&(k<n))
      {
        j=i;
        while (j<n)
          {
            printf("%4.d",x[j]);
            j=j+m;
            k++;
          }
        printf("\n");
        i++;
      }

  }

4 3
1 2 30 40

For this input there is no solution!!
No way exists for 4 numbers to be placed into 3 columns, when 2 columns has equal length and 3rd has littler length!
(but this input doesn't permitted by the statement of the task o_O)

alwasy
->
always

program t1506;
var
  n,m,a,b,c,i,j,l:integer;
  k,o:real;s:string;
  h:array [1..100] of integer;
  q:array [1..10,1..10] of integer;
  p:array [1..10,1..10] of integer;
begin
  readln(n,k);
  for i:=1 to n do
  read(h[i]);
  m:=trunc(sqrt(n));
  if frac(sqrt(n))<>0 then inc(m);
  for i:=1 to m do
  for j:=1 to m do
  begin
    inc(b);
    p[i,j]:=h[b];
    str(h[b],s);
    q[i,j]:=length(s);
    if q[i,j]>a then a:=q[i,j];
  end;
  for i:=1 to m do
  for j:=1 to m do
  begin
      write(' ');
      for b:=1 to (a-q[j,i]) do write(' ');
    if j<>m then
    begin
      write(p[j,i]);
      inc(l);
    end
    else
    begin
      writeln(p[j,i]);
      inc(l);
    end;
    if l=n then halt;
  end;
end.

Edited by author 03.03.2010 10:22

which answer?
n=11,k=5, '1 2 3 4 5 6 7 8 9 10 11';
1 4 7 10 ?
2 5 8 11 ?
3 6 9 ? ?
or
1 4 6 8 10
2 5 7 9 11
3 ? ? ?

My answer is right on computer's test, but I don't understand, why I've got WA#1.
Please, help me.
Here my code

# include<iostream.h>
int main()
{
    int n, k, a[101], i, j=0, h=0, p, b[101][101];
    cin>>n>>k;
    if (n%k==0)
        p=n/k;
    else
        p=n/k+1;
    for (i=0; i<n; i++)
    {
        cin>>a[i];
        if (j==p)
        {
            j=0;
            b[j++][++h]=a[i];
        }
        else
            b[j++][h]=a[i];
    }
    for (i=0; i<p; i++)
    {
        for (j=0; j<k && b[i][j]>=0 && b[i][j]<1000;j++)
        {
            if (b[i][j]>=0 && b[i][j]<=9)
                cout<<"   "<<b[i][j];
            else if (b[i][j]>=10 && b[i][j]<=99)
                cout<<"  "<<b[i][j];
            else
                cout<<" "<<b[i][j];
        }
        if (i<p-1)
            cout<<endl;
    }
    return 0;
}

Edited by author 03.06.2008 20:40

Is this Test correct?
4 3
1 2 3 4

>>
   1   3
   2   4

my AC prog. gives me that answer because

1 2 3
___4
isn't correct 100%.
{same test 16 5}

Edited by author 09.04.2008 23:39

Although my program passes all tests indicated in webboard, I have WA#3... My algo is quite simple. I increase the number of rows from one until it fits in K columns... Where is my fault?

Thx in advance

what is the test? Say please
I check all tests but all is correct what is worng?