C语言通过双向循环链表解决Josephus(约瑟夫)问题

题目描述：
n个人围坐一圈，标号1-n，从s开始报数，第m个报的人出列，一直循环下去直到所有人出列。设计一算法，输入n，m，s，输出出列顺序。

这个问题有好多种算法，这里用双向循环链表实现的

/* *use two-directioned looped linkedList */#include <cstdio>using namespace std ;class node{public:    node* last ;    node* next ;    int element ;    node( )    {        last = next = NULL ;        element = 0 ;    }    node( int element )    {         last = next = NULL ;         this->element = element ;    }} ;typedef node* node_pointer ;//pick the list recommended outvoid func( int n , int m , int s ){    //there is only one element in the list    if( n == 1 )    {        printf( "1/n" ) ;    }    //there is only two element in the list    else if( n == 2 )    {        if( m & 1 )        {            printf( "%d %d/n" , s , 3 - s ) ;        }        else        {            printf( "%d %d/n" , 3 - s , s ) ;        }    }    else    {        //size:the size of the list        //cnt:the counter to find the start pointer        int size = n , cnt = 1 ;        node_pointer head = new node( 1 ) ;//the first node pointer built        node_pointer current = head , start_pointer ; //start_pointer:the start pointer        if( s == 1 )        {            start_pointer = current ;        }        for( int i = 2 ; i <= size ; i ++ )        {            node_pointer tmp = new node( i ) ;            current->next = tmp ;            tmp->last = current ;            current = tmp ;            if( ++ cnt == s )            {                start_pointer = current ;            }        }        current->next = head ;        head->last = current ;                current = start_pointer ;        cnt = 1 ;        //the loop to pick the list out        while( true )        {            if( cnt == m )            {                if( size  >= 3 )                {                    printf( "%d " , current->element ) ;                    node_pointer tmp1 = current->last ;                    node_pointer tmp2 = current->next ;                    tmp1->next = tmp2 ;                    tmp2->last = tmp1 ;                                        delete current ;                    current = tmp2 ;                    cnt = 1 ;                                        size -- ;                }                else if( size == 2 )                {                    printf( "%d " , current->element ) ;                    node_pointer tmp = current->next ;                    tmp->last = tmp->next = tmp ;                                        delete current ;                    current = tmp ;                    cnt = 1 ;                                        size -- ;                }                else if( size == 1 )                {                    printf( "%d/n" , current->element ) ;                    delete current ;                    break ; //over                }            }            else            {                current = current->next ;                cnt ++ ;            }        }    }}int main(){    int n , m , s ;//n:the total number , m:every m is picked , s:the start    while( scanf( "%d%d%d" , &n , &m , &s ) == 3 )    {        func( n , m , s ) ;    }    return 0 ;}

结果：