in my accepted program I checked if i <= 10000 but printed i + k, so for test like 6 0 2 it printed 10001 but I think correct answer should be 0

Good problem, thanks to author!

4 5 1
196

6 10 1
4656

15 27 4
0

2 4 1
40

2 4 2
24

2 4 3
24

2 4 20
24

2 4 4
30

2 4 5
30

23 24 5
0

1 4 6
385

3 4 50
66

Edited by author 13.06.2014 14:08

Oh... He actually uses all of his tiles. Sorry, I read statement inattentively.

deleted

Edited by author 07.10.2013 22:24

# 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");
}

Why WA3?

#include<stdio.h>

int val(int n)
{
    int i,count=0;
    if(n<=0)
    return n;
    else
    {
    for(i=1;i<=(int)sqrt(n);i++)
    {
       if(n%i==0)
       count++;
    }
    return count;
    }
}

int main()
{
    int flag,m,n,k,i;
    flag=0;
    scanf("%d %d %d",&m,&n,&k);
    for(i=1;i<=10000;i++)
    {
        if(val(i)==n&&val(i-k)==m)
        {
             flag=1;
             break;
        }
    }
    if(flag==1)
    printf("%d",i);
    else
    printf("0");
    //system("pause");
    return 0;
}


I am Getting Wrong Answer For Test Case 1 Itself

But i am getting the right answer 16 For The Input Cases

You just need to check all possible numbers of L (1..10000) to solve this. So you get O(n^2) solution.

who can explain the problem?

var
 a:array[1..10000]of integer;
 m,n,k,i,j:integer;

begin
 fillchar(a,sizeof(a),0);
 readln(m,n,k);
 for i:=k+1 to 10000 do
  begin
   for j:=1 to trunc(sqrt(i)) do
    if i mod j=0 then a[i]:=a[i]+1;
   if (a[i]=n)and(a[i-k]=m) then begin writeln(i);halt;end;
  end;
end.

#include<stdio.h>
#include<math.h>
int yue(int n)
{
    int p=1,i;
    for(i=2;i<=(int)sqrt(n);i++)
        if(n%i==0)
        p++;
    return p;
}
int main ()
{
    int m,n,k,i,test;
    while(scanf("%d %d %d",&m,&n,&k)!=EOF)
    {
        test=0;
        for(i=1;i<=1000;i++)
        if(n==yue(i))
        {
            if(m==yue(i-k))
            {
                 printf("%d\n",i);
                 test=1;
                 break;
            }
        }
        if(test==0)
            printf("0\n");


    }
    return 0;

}

[code deleted]

Edited by moderator 28.05.2007 17:49

If text were like this (You are given three numbers – M, N and K. You should find the less number L, such as you can form !!!EXACTLY!!! N different rectangles using all EXACTLY L tiles, and form M rectangles using L-K tiles.),  I would solve this problem whithout so many submittions=)

P.S. Yo shouldn't correct text, this is my whish =)

Amount of rectangles is equal to the number of divisiors of tiles count wich are no more than Square_Root(tiles count). So the part of code solving this problem may be the following:

  For L:=1 To 10000-K Do Begin
    DivisiorsCount:=1;
    For J:=2 To Trunc(Sqrt(L)) Do If L Mod J=0 Then
      Inc(DivisiorsCount);

    If DivisiorsCount=M Then Begin
      DivisiorsCount2:=1;
      For J:=2 To Trunc(Sqrt(L+K)) Do If (L+K) Mod J=0 Then
        Inc(DivisiorsCount2);

      If DivisiorsCount2=N Then Begin
        SoulutionExists:=True;
        Break;
      End;
    End;
  End;

You should output L+K if SolutionExists, 0 - otherwise.

program windy;
    var
      n,i,m,k,j,n1,n2:longint;
      c:boolean;
    begin
      readln(m,n,k);
      c:=true;
      i:=k+1;
      while c and (i<=10000) do
        begin
          n1:=0;
          n2:=0;
          for j:=2 to trunc(sqrt(i)) do
            begin
              if i mod j=0 then n1:=n1+1;
              if (i-k) mod j=0 then n2:=n2+1;
            end;
          if (n1+1=n) and (n2+1=m) then c:=false
            else i:=i+1;
         end;
       if c then writeln(0);
       if not c then writeln(i)
     end.

My program is here:

var
   m,n,k:longint;
   min:longint;
function nums(a:longint):longint;
var
   i,tot:longint;
begin
     tot:=0;
     for i:=1 to trunc(sqrt(a)) do
     if a mod i=0 then inc(tot);
     nums:=tot;
end;
begin
     readln(m,n,k);
     for min:=1 to 10000 do
     begin
          if (nums(min)=n)and(nums(min-k)=m) then
          begin
               writeln(min);
               exit;
          end;
     end;
     writeln(0);
end.