SPOJ.com - Problem RGB7311

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RGB7311 - Гуравын зэрэгтэд хураа

Өгөгдсөн тоог 3-ын зэрэгтэд хураа.

Input

Бүхэл тоо өгөгдөнө.

Output

Хураагдсан тоо.

Example

Input:

36 Output:

4

Submit solution!

Нэмсэн:	Bataa
Огноо:	2013-01-11
Хугацааны хязгаарлалт:	1s
Эх кодын хэмжээний хязгаарлалт:	50000B
Memory limit:	1536MB
Cluster:	Cube (Intel G860)
Програмчлалын хэлүүд:	ADA95 ASM32 BASH BF C NCSHARP CSHARP C++ 4.3.2 CPP C99 CLPS LISP sbcl LISP clisp D ERL FORTRAN HASK ICON ICK JAVA JS-RHINO JULIA LUA NEM NICE OCAML PAS-GPC PAS-FPC PERL PHP PIKE PRLG-swi PYTHON PYPY3 PYTHON3 RUBY SCALA SCM guile ST TCL WHITESPACE

hide comments

2019-10-11 08:49:20
oorsdoo huugeech orc uudaa

2019-10-09 13:59:22
GG
LOL
OOFFFF

2019-04-09 15:42:10

#include <iostream>
#include<math.h>
using namespace std;
int main()
{
int n;
scanf("%d",&n);
while(n%3==0)
{
n=n/3;
}
printf("%d",n);
}

Last edit: 2019-04-09 15:44:45

2019-03-12 04:01:07
zow bodod bhad buruu ged bh ymaaa

2019-03-01 10:17:25
#include <iostream>
#include<math.h>
using namespace std;
main()
{ long long n, r, a, i = 0;
cin>>n;
a=n;
r=pow(3,i);
while (n>=r) {
if (n%r==0) {
a=n/r;
}
i=i+1;
r=pow(3,i);
}
cout<<a<<endl;
return 0;
}

2019-02-09 07:19:37
#include <iostream>
#include<math.h>

using namespace std;

int main()
{
int n, r, a, i = 0;
cin >> n;
a = n;
r = pow( 3, i );
while ( n >= r ) {
if ( n % r == 0) {
a = n / r;
}

i = i + 1;

r = pow( 3, i );
}
cout << a << endl;

return 0;
}
//hN

Last edit: 2019-02-09 07:20:16

2018-12-13 02:49:19
#include<iostream>
using namespace std;
int main(){
int a,b=1,c,i;
cin >> a;
int l=a;
while(c%b==0){
b=b*3;
c=a/b;
}
cout << c;
return 0;
}

zow l baih bodson ym shig baihiin buruu gd bhiin medqeeee

2018-11-15 07:19:43
zuv bolood bhd buruu hariult gd bhin.
eniigee zasaj uguuchee SPOJ iihan.
hezee zashiin bee?

АЛГОРИТМЫН ЦАГААН ТОЛГОЙ

RGB7311 - Гуравын зэрэгтэд хураа

Input

Output

Example

Input:

36

Output:

4