11773 - King's Wish

[nealzane]

for the example input 18, how is an output '9 1' violating the problem statement?
it seems '9 1' is better than '9 2'
can '9 1' cover a square smaller than 18x18?

[emotional blind]

can '9 1' cover a square smaller than 18x18?

Yes, it can cover, 9 x 9

[nealzane]

can '9 1' cover a square smaller than 18x18?

Yes, it can cover, 9 x 9

right, i was aware of it. but for '9 1' i guess 9x9 is not the problem wants because '9 1' covers a length of the square.

the problem says '2 1' is valid for 4x4. while if '9 1' is valid for 9x9, '2 1' cannot be valid for 4x4, since in that case, '2 1' is valid for 2x2 also which is smaller than 4x4.

i guess '9 1' cannot be used for 9x9, just as '2 1' cannot be used for 2x2.

[orfest]

Guys, could you please give any difficult test cases?

[orfest]

Is 3 20 correct answer for test case 60?

[emotional blind]

can '9 1' cover a square smaller than 18x18?

Yes, it can cover, 9 x 9

right, i was aware of it. but for '9 1' i guess 9x9 is not the problem wants because '9 1' covers a length of the square.

the problem says '2 1' is valid for 4x4. while if '9 1' is valid for 9x9, '2 1' cannot be valid for 4x4, since in that case, '2 1' is valid for 2x2 also which is smaller than 4x4.

i guess '9 1' cannot be used for 9x9, just as '2 1' cannot be used for 2x2.

I think you did not understand the problem properly. In problem statement "For example, if K = 6, you cannot use (2, 1) tile as you can fill 4x4 square with it. (The king does not like smaller replica of his room with same size of tile)." - this line wants to say that 4 x 4 is a smaller possible square, but this is not smallest. The smallest possible square using (2, 1) is 2 x 2.

[emotional blind]

Is 3 20 correct answer for test case 60?

No, I think right answer is 4 30

[nealzane]

thanks Arif for the explanation.

[md_yusuf]

i have not find any simple equation for this problem im getting tle. plz give me a hints. here is my code....


#include<iostream>
using namespace std;
int main()
{
long int n,flag,i,x,j,l,w;
scanf("%ld",&n);
for(i=1;i<=n;i++)
{
flag=0;
scanf("%ld",&x);
for(j=2;j*j<=x;j++)
{
if((x%j)==0)
{
w=j;
l=x/j;
if((l%w)!=0)
{
flag=1;
break;
}
}
}
if(flag==0)
printf("Case %ld: Impossible\n",i);
else
printf("Case %ld: %ld %ld\n",i,l,w);
}
return 0;
}

[md_yusuf]

I think my algo is wrong ....
nobody help me