[Microsoft面试]题目：n 个数字（0,1,…,n-1）形成一个圆圈，从数字0 开始，每次从这个圆圈中删除第m 个数字（第一个为当前数字本身，第二个为当前数字的下一个数字）。当一个数字删除后，从被删除数字的下一个继续删除第m 个数字。求出在这个圆圈中剩下的最后一个数字。

Actually,although this is a so traditional problem, I was always to lazy to think aboutthis or even to search for the answer.(What a shame...). Finally, by google Ifound the elegant solution for it.The keys are:1) if we shift the ids by k, namely, start fromk instead of 0, we should add the result by k%n2) after the first round, we start from k+1 (possibly % n) with n-1 elements, that is equal to an (n-1) problem while startfrom (k+1)th element instead of 0, so the answer is (f(n-1, m)+k+1)%n3) k = m-1, so f(n,m)=(f(n-1,m)+m)%n.finally, f(1, m) = 0;Now this is a O(n) solution.int joseph(int n, int m) {  int fn=0;  for (int i=2; i<=n; i++) {    fn = (fn+m)%i;  }  return fn;}hu...长出一口气。。。