Solved

Help with algorithm

Posted on 2000-02-13
2
211 Views
Last Modified: 2006-11-17
I got this algorithm for a function that solves Josephus algorithm. and I will like to translate it to c++.
In C langiage:

#include <stdio.h>
#include <stdlib.h>

int josef(int num, int skip)
{
char *vec;
int i;
int curPtr;
int numLive = num;

vec = (char *)malloc(num+1);
for(i=1 ; i<=num ; i++)
   vec[i] = 1;

curPtr = 1;
while(numLive > 1)
{
   i = 1;
   while(i<skip)
   {
      curPtr++;
      if(curPtr > num)
         curPtr = 1;

      if(vec[curPtr])
         i++;
   }

   vec[curPtr] = 0;  /* KILL him*/

   curPtr++;
   if(curPtr > num)
      curPtr = 1;


   numLive--;
}

/* Find the one LIVE */
for(i=1 ; i<=num ; i++)
   if(vec[i])
      return(i);

return(-1);
}

What I have done so far in C++:

Jos(n,m)
for i=0 to n
  {a[i]=i+1;}
while n>i  do
 for i=0 to n
 {
  if( a[i]!=0)
  {
   count=count+1;
   if (count==m)
   {
     a[i]=0;
     count=0;
    }
   }
  }

I'm I doing it okay?
what else do I need?
0
Comment
Question by:milalik
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
2 Comments
 

Author Comment

by:milalik
ID: 2517044
josephus problem is related to the following. Begining with n people arranged in a circle. Starting with the first person, count continuosly around the circle removing every m-th person until only one remains.Example n=7 and m=3
         1,2,3,4,5,6,7
          1,2,4,5,6,7
           1,2,4,5,7
            1,4,5,7
             1,4,5
              1,4
               4
0
 
LVL 22

Accepted Solution

by:
nietod earned 100 total points
ID: 2517076
The C code _IS_ C++ code.  So you can just use that.  however it can be made a little more "C++ish"  (Also it has a memory leak that this fixes.)

#include <stdio.h>
#include <stdlib.h>

int josef(int num, int skip)
{
   char *vec = new char[num + 1];
    int numLive = num;

   for(int i=1 ; i<=num ; i++)
      vec[i] = 1;

   int curPtr = 1;
   while(numLive > 1)
   {
      int i = 1;
      while(i<skip)
      {
         curPtr++;
         if(curPtr > num)
            curPtr = 1;
         if(vec[curPtr])
            i++;
       }

      vec[curPtr] = 0;  /* KILL him*/
      curPtr++;
      if(curPtr > num)
         curPtr = 1;
      numLive--;
   }

   int RetVal = -1;

   /* Find the one LIVE */
   for(int i=1 ; i<=num ; i++)
      if(vec[i])
      {
          RetVal = i;
         break;
      }

    delete [] vec;
    return (RetVal);
}
0

Featured Post

Technology Partners: We Want Your Opinion!

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Often, when implementing a feature, you won't know how certain events should be handled at the point where they occur and you'd rather defer to the user of your function or class. For example, a XML parser will extract a tag from the source code, wh…
Container Orchestration platforms empower organizations to scale their apps at an exceptional rate. This is the reason numerous innovation-driven companies are moving apps to an appropriated datacenter wide platform that empowers them to scale at a …
The goal of the video will be to teach the user the concept of local variables and scope. An example of a locally defined variable will be given as well as an explanation of what scope is in C++. The local variable and concept of scope will be relat…
The viewer will learn how to pass data into a function in C++. This is one step further in using functions. Instead of only printing text onto the console, the function will be able to perform calculations with argumentents given by the user.

632 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question