Josephus問題的一種解法（鏈表實現）

這裏並不是原始的Josephus問題，而是其“通用版”，而歷史上的那個只是個特殊情況
下面直接給代碼:

#include <stdio.h>
#include <stdlib.h>
struct People;
typedef struct People *PtrToPeople;
struct People
{
    int Position;
    PtrToPeople Next;
};
typedef PtrToPeople Curcle;
typedef PtrToPeople Position;
void Insert(Position P,int e)
{
    Position TmpCell;
    TmpCell = malloc(sizeof(struct People));
    TmpCell->Position = e;
    TmpCell->Next = P->Next;
    P->Next = TmpCell;
}
Position FindPre(Position P)
{
    Position T = P;
    for(;;)
        if(T->Next == P)
            return T;
        else
            T = T->Next;
}
Position DeleteAndPrint(Position P)
{
    printf("%d ",P->Position);
    Position T = FindPre(P);
    T->Next = P->Next;
    free(P);
    return T->Next;
}
/* 輸出被選中的，剩下最後一個是winner */
int main()
{
    int N,M,i;
    /* N表示總數，M表示跳格數 */
    scanf("%d%d",&M,&N);

    Curcle C = malloc(sizeof(struct People));
    C->Position = 1;
    C->Next = C;
    for(i = 2; i <= N; i++)
    {
        Insert(C,i);
        C = C->Next;
    }
    Position P = C->Next;
    for(;;)
    {
        int j;
        for(j = 0; j < M; j++)
        {
            P = P->Next;
        }
        if( P->Next == P )
        {
            break;
        }
        else
            P = DeleteAndPrint(P);
    }
    return 0;
}