Power Crisis

Problem Description
During the power crisis in New Zealand this winter (caused by a shortage of rain and hence low levels in the hydro dams), a contingency scheme was developed to turn off the power to areas of the country in a systematic,
totally fair, manner. The country was divided up into N regions (Auckland was region number 1, and Wellington number 13). A number, m, would be picked `at random', and the power would first be turned off in region 1
(clearly the fairest starting point) and then in every m'th region after that, wrapping around to 1 after N, and ignoring regions already turned off. For example, if N = 17 and m = 5, power would be turned off to the
regions in the order:1,6,11,16,5,12,2,9,17,10,4,15,14,3,8,13,7. The problem is that it is clearly fairest to turn off Wellington last (after all, that is where the Electricity headquarters are), so for a given
N, the `random' number m needs to be carefully chosen so that region 13 is the last region selected. 
Write a program that will read in the number of regions and then determine
the smallest number m that will ensure that Wellington (region 13) can function while the rest of the country is blacked out.
Input
Input will consist of a series of lines, each line containing the number of regions (N) with . The file will be terminated by a line consisting of a single 0.
Output
Output will consist of a series of lines, one for each line of the input. Each line will consist of the number m according to the above scheme.
Sample Input

17
0

Sample Output

7

 
   
 //       ：    N,  M      ，      13， M  。
  
 
   
//  ：
  
#include
#include
#include
using namespace std;
int main()
{
	//freopen("a.txt","r",stdin);
    list l;
    list ::iterator it1,it2,it3;
    int i,j,k,n;
    while(scanf("%d",&n),n)
    {
      for(i=1;;i++)
      {   l.clear();
          k=0;
          for(j=1;j<=n;j++) l.push_back(j); 
          it1=l.begin();
          while(l.size()>1)
          {  it2=it1; it1++;
            if(it1==l.end()) it1=l.begin();
            l.erase(it2);
            k=(i-1)%l.size();
            while(k--)
            {    it1++;
                if(it1==l.end()) it1=l.begin();
            }
          }
         if(l.front()==13) {cout<