The Joseph's problem is notoriously known. For those who are not familiar with the original problem: from among n people, numbered 1, 2, ..., n, standing in circle every mth is going to be executed and only the life of the last remaining person will be saved. Joseph was smart enough to choose the position of the last remaining person, thus saving his life to give us the message about the incident. For example when n = 6 and m = 5 then the people will be executed in the order 5, 4, 6, 2, 3 and 1 will be saved.
Suppose that there are k good guys and k bad guys. In the circle the first k are good guys and the last k bad guys. You have to determine such minimal m that all the bad guys will be executed before the first good guy.
設 ai 是第i次踢掉的人在第i-1次踢掉後剩下的人中是第幾個。那麼
a(n) = [a(n-1)+m-1]mod(2k-n+1)
要求a(n) > k;n = 1,2,3,...,k
其中2k-n+1是第i-1次踢人後剩下的人數。
可以設計如下算法:
{
int n;
for(n=1;n<=k;n++)
{
a = (a+m-1)%(k2-n+1);
if(a == 0) a = k2-n+1;
if(a<=k && a>=1) return false;
}
return true;
}
然後,我們注意到,第一次踢的人是 m%2k,我們要求 m%2k > k,也就是 m = 2k*r+h,h>k,那麼就可以設計如下的算法找出最小的m:
{
for(h=k+1;h<=2*k;h++)
{
m = 2*k*r+h;
if(Joseph(k,m)) goto end; // 找到m跳出
}
}
end: