"Algorithm" Joseph Ring C + + source code

#include <iostream>using namespacestd;intMain () {intA= -, b=3, winner=0;//Total number A, number to B in multiples of left, last man winner     for(intI=2; i<=a;i++) Winner= (winner+b)%i; cout<<"Winner:"<<winner+1<<Endl;}

We consider the following process

1 2 3 4 5 6 ..... K-1 k k+1 ... n-1 n

The first number for the K-brother hangs, and then the rest of the n-1 individuals, from the k+1 continue.

K+1 k+1 .... n-1 N 1 2 3 4 5 6,

The sequence numbers are all minus k, resulting in the following sequences:

1 2 3 4 5 6 ..... N-1

That is n-1 personal situation.

Assuming that the last remaining person, in the sequence of the first (n-1) person is f (n-1), then he is in the N-man sequence, numbered (K+f (n-1))%n, and has our recursive formula:

F (n) = (k+f (n-1))%n;

F (1) = 1;

http://blog.csdn.net/ice110956/article/details/12906065

"Algorithm" Joseph Ring C + + source code