Posted: Mon Jan 24, 2005 8:41 pm
i don't understand the problem ...this is my wrong!!THx OMega
i don't understand the problem ...this is my wrong!!THx OMega
Page 8 of 11
i don't understand the problem ...this is my wrong!!THx OMega
Posted: Mon Jan 24, 2005 10:59 pm
Hi PerHagen:
In this problem you must find the 1500th ugly number and print it.
Ugly numbers are numbers which are only divisible by 2, 3, 5, that is 2^a*3^b*5^c.
Did you get it??? Read the problem carefully and you will understand.
Hope it helps...
Yandry.
In this problem you must find the 1500th ugly number and print it.
Ugly numbers are numbers which are only divisible by 2, 3, 5, that is 2^a*3^b*5^c.
Did you get it??? Read the problem carefully and you will understand.
Hope it helps...
Yandry.
136-why PE
Posted: Sun Apr 10, 2005 4:27 pm
#include<stdio.h>
int main(void)
{unsigned long n = 1500 , un=********;
printf("The %lu'th ugly number is %lu",n,un);
return 0;}
int main(void)
{unsigned long n = 1500 , un=********;
printf("The %lu'th ugly number is %lu",n,un);
return 0;}
Posted: Sun Apr 10, 2005 4:44 pm
Hi
use this
printf("The %lu'th ugly number is %lu\n",n,un);
instead of this
printf("The %lu'th ugly number is %lu",n,un);
MAP
use this
printf("The %lu'th ugly number is %lu\n",n,un);
instead of this
printf("The %lu'th ugly number is %lu",n,un);
MAP
Posted: Sun Apr 10, 2005 4:46 pm
P.E. stands for Presentation Error. It happens when your solution doesn't conform to output specification, but apart from that the result is fine.
In this case you're obviously missing an end of line at the end of the output.
In this case you're obviously missing an end of line at the end of the output.
136 - 4e za prikol
Posted: Thu Apr 28, 2005 7:02 am
what is wrong with this function
i want to find how many ugly numbers less or equal to 'x'...
Function F(x : LongInt) : LongInt;
var x2,x3,x5 : LongInt;
cnt,T : Int64;
i,j,k : LongInt;
Begin
x2 := trunc(Ln(x)/Ln(2));
x3 := trunc(Ln(x)/Ln(3));
x5 := trunc(Ln(x)/Ln(5));
cnt := 0;
For i := 0 to x2 do
For j := 0 to x3 do
For k := 0 to x5 do Begin
T := int64(trunc(exp(Ln(2)*i)));
T := T*int64(trunc(exp(Ln(3)*j)));
T := T*int64(trunc(exp(Ln(5)*k)));
if T <= x then cnt := cnt+1;
end;
F := cnt;
end;
P136 WA. Why??
Posted: Mon May 02, 2005 8:09 am
Hi.
My solution on P136 gets WA.
Can anyone help me?
Here is my solution:
My solution on P136 gets WA.
Can anyone help me?
Here is my solution:
Code: Select all
#include <stdio.h>
#include <stdlib.h>
double K, A [1600];
double Min (double A, double B) { return A < B ? A : B; }
int main (void) {
/* freopen ("output.txt", "w", stdout); */
long I, J;
double M;
A [0] = 1;
printf ("The 1500'th ugly number is");
for (I = 0; I < 1500; I++) {
M = A [I];
K = 2000000000;
for (J = I; J >= 0 && 5*A [J] > M; J--) {
if (2*A [J] > M) K = Min (K, 2*A [J]);
if (3*A [J] > M) K = Min (K, 3*A [J]);
K = Min (K, 5*A [J]);
}
A [I+1] = K;
printf (" %.0f", A [I]);
}
printf (".\n");
exit (0);
} Posted: Mon May 02, 2005 8:23 am
I accepted it. I thought that I must print all ugly numbers with index from 1 to 1500 
.
.Posted: Mon May 02, 2005 6:14 pm
Post here what this function is doing 
.
.Posted: Tue May 03, 2005 8:36 am
Rezult of F(x) is equal to count of numbers
that are ugly and less or equal to x.
that are ugly and less or equal to x.
Posted: Tue May 03, 2005 8:27 pm
To solve this problem use DP.
I solved it with this algor.:
Let first K ugly number is U [1] ... U [K].
For all I (1..K) M = Min3 (2*U , 3*U , 5*U )
and U [K+1] = Min (M (I = 1..K)).
GL.
I solved it with this algor.:
Let first K ugly number is U [1] ... U [K].
For all I (1..K) M = Min3 (2*U , 3*U , 5*U )
and U [K+1] = Min (M (I = 1..K)).
GL
.Posted: Wed May 04, 2005 8:35 am
Thank you, but this one works O(1500^2) isn't it. What is the time you got on AC for this task, could you tell me? I wanted to find that with Binary search using F(x). And I did it!!! Thank you. I even didn't think about DP for this question. And I will be able to solve this like question using your method. Thank you for New Ideas.
GL.
GL.
Posted: Wed May 04, 2005 4:59 pm
I solved it and it didn't got TL.
hm.
Posted: Thu Jun 23, 2005 12:31 am
( 2 * x )
or
( 3 * y )
or
( 4 * z )
<= 1500
prepare.
x = 1, y = 1, z = 1 and the STACK[0] = 1
first.
STACK[1] number is smallest number in ( 2*x ), ( 3*y), ( 4*z ) .
you can found it ( 2 * x ) when x = 1 ; it is smallest number ;
so, STACK[1] = 2 .
and next number is must be larger then 2
so, set x=2, y=1 and z=1.
x=2, y=1, z=1
the next number is y=1 ( 3 ) -> STACK[3] , y= 2.
.....
1
2
3
4
6
...
......
or
( 3 * y )
or
( 4 * z )
<= 1500
prepare.
x = 1, y = 1, z = 1 and the STACK[0] = 1
first.
STACK[1] number is smallest number in ( 2*x ), ( 3*y), ( 4*z ) .
you can found it ( 2 * x ) when x = 1 ; it is smallest number ;
so, STACK[1] = 2 .
and next number is must be larger then 2
so, set x=2, y=1 and z=1.
x=2, y=1, z=1
the next number is y=1 ( 3 ) -> STACK[3] , y= 2.
.....
1
2
3
4
6
...
......
Posted: Wed Jul 06, 2005 10:03 pm
I thought the same thing so what are we supposed to do