F - System Overload（约瑟夫环变形）

Description

Recently you must have experienced that when too many people use the BBS simultaneously, the net becomes very, very slow.
To put an end to this problem, the Sysop has developed a contingency scheme for times of peak load to cut off net access for some buildings of the university in a systematic, totally fair manner. Our university buildings were enumerated randomly from 1 to n. XWB is number 1, CaoGuangBiao (CGB) Building is number 2, and so on in a purely random order.
Then a number m would be picked at random, and BBS access would first be cut off in building 1 (clearly the fairest starting point) and then in every mth building after that, wrapping around to 1 after n, and ignoring buildings already cut off. For example, if n=17 and m=5, net access would be cut off to the buildings 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 cut off CGB Building last (after all, this is where the best programmers come from), so for a given n, the random number m needs to be carefully chosen so that building 2 is the last building selected.

Your job is to write a program that will read in a number of buildings n and then determine the smallest integer m that will ensure that our CGB Building can surf the net while the rest of the university is cut off.

Input Specification

The input file will contain one or more lines, each line containing one integer nwith 3 <= n < 150, representing the number of buildings in the university.
Input is terminated by a value of zero (0) for n.

Output Specification

For each line of the input, print one line containing the integer m fulfilling the requirement specified above.

Sample Input

Sample Output

2

5

2

4

3

11

2

3

8

16
解题思路：一开始第1层楼就已被断电，要求选出步长k使得每累加k层楼循环进行断电，要求最后剩下的一层楼是第2层楼。回顾一下约瑟夫环问题：已知n个人（0~n-1）围坐在一张圆桌周围。从编号0的人开始报数，数到k-1的那个人出列；他的下一个人又从0开始报数，数到k-1的那个人又出列；依此规律重复下去，直到圆桌周围的人全部出列。而此题中第1个人（编号为0）就已经出列了，我们按递推式f[n]=(f[n-1]+k)%n计算出第n个人即最后一个人出列的编号f[n]，这个一开始是第k-1个人出列，所以我们只需将原来的0~n-1重新编号，偏移量是1-k，标记第0个人在第k-1个位置（表示第一次就出列），则最终出列的编号为((f[n]+1-k)%n+n)%n（防止编号出现负数或者是超过n）。
AC代码：

 #include<iostream>

 #include<cstdio>

 using namespace std;

 int main(){

     int n,s;

     while(cin>>n&&n){

         for(int k=;;++k){//如果是1的话，第2层楼一开始就会被断电，因此步长从2开始

             s=;

             for(int j=;j<=n;++j)s=(s+k)%j;

             s=((-k+s)%n+n)%n;

             if(s==){cout<<k<<endl;break;}//第二层楼的编号为1，最后一个断网的楼层是2，即找到最小的步长k，直接退出

         }

     }

     return ;

 }