2013
11-10

Three powers

Consider the set of all non-negative integer powers of 3.

S = { 1, 3, 9, 27, 81, … }

Consider the sequence of all subsets of S ordered by the value of the sum of their elements. The question is simple: find the set at the n-th position in the sequence and print it in increasing order of its elements.

Each line of input contains a number n, which is a positive integer with no more than 19 digits. The last line of input contains 0 and it should not be processed.

For each line of input, output a single line displaying the n-th set as described above, in the format used in the sample output.

1
7
14
783
1125900981634049
0


{ }
{ 3, 9 }
{ 1, 9, 27 }
{ 3, 9, 27, 6561, 19683 }
{ 59049, 3486784401, 205891132094649, 717897987691852588770249 }


//* @author:
import java.math.BigInteger;
import java.util.Scanner;

public class Main {
public static void main(String[]args){
BigInteger dat[]=new BigInteger[100];
dat[0]=BigInteger.ONE;
BigInteger tr=new BigInteger("3");
for(int i=1;i< 100;i++){
dat[i]=dat[i-1].multiply(tr);
}
long N;
int t;
boolean first;
Scanner cin=new Scanner(System.in);
while(cin.hasNext()){
N=cin.nextLong();
if(N==0) break;
N--;
t=0;
first=true;
System.out.print("{");
while(N>0){
if(N%2==1){
if(first){
first=false;
System.out.print(" "+dat[t]);
}
else System.out.print(", "+dat[t]);
}
t++;
N/=2;
}
System.out.println(" }");
}
}
}

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3. /*
* =====================================================================================
*
* Filename: 1366.cc
*
* Description:
*
* Version: 1.0
* Created: 2014年01月06日 14时52分14秒
* Revision: none
* Compiler: gcc
*
* Author: Wenxian Ni (Hello World~), [email protected]
* Organization: AMS/ICT
*
* =====================================================================================
*/

#include
#include

using namespace std;

int main()
{
stack st;
int n,i,j;
int test;
int a[100001];
int b[100001];
while(cin>>n)
{
for(i=1;i>a[i];
for(i=1;i>b[i];
//st.clear();
while(!st.empty())
st.pop();
i = 1;
j = 1;

while(in)
break;
}
while(!st.empty()&&st.top()==b[j])
{
st.pop();
j++;
}
}
if(st.empty())
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}
return 0;
}