关于Joesphus问题

[VC_Mio]

有N人围成一圈，编号为1至N，从1号开始传递一物体，M次传递之后，持有物体的人退出，圈缩小，从退出的下一人开始，继续进行.....最后留下来到人为？

下面是书上面的代码，哪位高手能解释一下么？
PS：我自己写的也能算出一样的结果，但是看不明白下面这种算法....


#include <iostream>
#include <list>
using namespace std;
int main()
{
int i,j, n, m, mPrime, numLeft;
list <int > L;
list<int>::iterator iter;
//Initialization
cout<<"enter N (# of people) & M (# of passes before elimination):";
cin>>n>>m;
numLeft = n;

mPrime = m % n;
for (i=1 ; i<= n; i++)
L.push_back(i);
iter = L.begin();
// Pass the potato
for (i= 0; i< n; i++)
{
mPrime = mPrime % numLeft;
if (mPrime <= numLeft/2) // pass forward
for (j = 0; j < mPrime; j++)
{
iter++;
if (iter == L.end())
iter = L.begin();
}
else // pass backward
for (j = 0; j < mPrime; j++)
{
if (iter == L.begin())
iter = --L.end();
else
iter--;
}
cout<<*iter<<" ";
iter= L.erase(iter);
if (iter == L.end())
iter = L.begin();
}
cout<<endl;
return 0;
}