用数据结构c++顺序表和链表实现约瑟夫环问题，输出每次出列的编号和幸存者的编号。

顺序表实现：

#include <stdio.h>

#define MAX_SIZE 100

typedef struct {
    int data[MAX_SIZE];
    int length;
} SqList;

int josephus(SqList *list, int k) {
    int i, index = 0;
    while (list->length > 1) {
        index = (index + k - 1) % list->length;
        printf("%d 出列\n", list->data[index]);
        for (i = index; i < list->length - 1; i++) {
            list->data[i] = list->data[i + 1];
        }
        list->length--;
    }
    return list->data[0];
}

int main() {
    SqList list = {{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, 10};
    int k = 3;
    int survivor = josephus(&list, k);
    printf("幸存者编号：%d\n", survivor);
    return 0;
}

链表实现：

#include <stdio.h>
#include <stdlib.h>

typedef struct Node {
    int data;
    struct Node *next;
} ListNode;

ListNode *createList(int n) {
    int i;
    ListNode *head, *p;
    head = (ListNode *) malloc(sizeof(ListNode));
    head->next = NULL;
    p = head;
    for (i = 1; i <= n; i++) {
        ListNode *node = (ListNode *) malloc(sizeof(ListNode));
        node->data = i;
        node->next = NULL;
        p->next = node;
        p = node;
    }
    p->next = head->next;
    return head;
}

int josephus(ListNode *head, int k) {
    int i, index = 0;
    ListNode *p = head, *q;
    while (p->next != p) {
        for (i = 1; i < k; i++) {
            q = p;
            p = p->next;
        }
        index = p->data;
        printf("%d 出列\n", index);
        q->next = p->next;
        free(p);
        p = q->next;
    }
    return p->data;
}

int main() {
    int n = 10, k = 3;
    ListNode *head = createList(n);
    int survivor = josephus(head, k);
    printf("幸存者编号：%d\n", survivor);
    return 0;
}