oj数组函数题，约瑟夫报数问题

Question

今天的题有点难，有能力的来挑战一下吧，（凑一下30字还没到30字么.）

Answer 1

void CountOff(int n, int m, int out[])
{
  int i, j, ct;
  int data[MAXN];
  for (i = 0; i < n; i++)
    data[i] = i + 1;
  i = j = ct = 0;
  while (ct < n - 1)
    {
      if (data[i] != 0)
        j++;
      if (j == m)
        {
          out[ct++] = data[i];
          data[i] = 0;
          j = 0;
        }
      i++;
      if (i == n)
        i = 0;
    }
  for (i = 0; i < n; i++)
    if (data[i] != 0)
      {
        out[ct] = data[i];
        break;
      }
}