poj 2244 Eeny Meeny Moo

 

Eeny Meeny Moo

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2150		Accepted: 1459

Description

Surely you have made the experience that when too many people use the Internet simultaneously, the net becomes very, very slow. 
To put an end to this problem, the University of Ulm has developed a contingency scheme for times of peak load to cut off net access for some cities of the country in a systematic, totally fair manner. Germany's cities were enumerated randomly from 1 to n. Freiburg was number 1, Ulm was number 2, Karlsruhe was number 3, and so on in a purely random order. 
Then a number m would be picked at random, and Internet access would first be cut off in city 1 (clearly the fairest starting point) and then in every mth city after that, wrapping around to 1 after n, and ignoring cities already cut off. For example, if n=17 and m=5, net access would be cut off to the cities 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 Ulm 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 city 2 is the last city selected. 

Your job is to write a program that will read in a number of cities n and then determine the smallest integer m that will ensure that Ulm can surf the net while the rest of the country is cut off.

Input

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

Output

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

Sample Input

3
4
5
6
7
8
9
10
11
12
0

Sample Output

2
5
2
4
3
11
2
3
8
16

Source

Ulm Local 1996

/× 令f(i,m)表示i个人玩游戏报m退出最后胜利者的编号，最后的结构就是f(n),递推公式为： f(i,m)=0,(i=1) f(i,m)=(f(i-1,m)+m)%i (i>1) 所以可以从1到n算出f的数值，最后结果就是f(n,m),注意生活中标号一般从1开始，所以输出时应输出f(n,m)+1. int main() { int n, m,i, s=0; printf( "N M = "); scanf("%d%d ",&n,&m); for(i=2;i<=n;i++)s=(s+m)%i; printf("The winner is %d/n ", s+1); } 是指n个人,编号从0到n-1 .输出的时候必须输出s+1 (编号s的人是第s+1个人) //第一个人总是最先被删除的。因此可以把该问题看成是n-1个人的问题, //希望最后问题中最原始的2留下来,找这个递增的m int cal(int n,int m) { int s=0; for(int i=2;i<=n-1;i++) s=(s+m)%i;//如果s从1开始，则(s+m-1)%i+1 return s; } int main() { int n; while(scanf("%d",&n)!=EOF) { if(n==0) break; for(int i=1;;i++) if(cal(n,i)==0) //编号为1的城市已经出列，所以第二座城市现在编号为1，但递推式的第一个值为0 { printf("%d/n",i); break; } } } ×/