运用循环单链表实现约瑟夫问题c语言代码

约瑟夫环是一个经典的算法问题，通常用于演示递归和模运算的概念。在循环单链表中实现约瑟夫问题，首先需要创建一个节点结构和一个头指针，然后模拟人们从列表中按顺序移除并跳过指定步骤的过程。

以下是一个简单的C语言代码示例：

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

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

Node* createNode(int value) {
    Node* newNode = (Node*)malloc(sizeof(Node));
    newNode->data = value;
    newNode->next = NULL;
    return newNode;
}

void josephusProblem(Node* head, int steps) {
    if (head == NULL || steps <= 0) {
        printf("Invalid input.\n");
        return;
    }

    Node* victim = head;
    for (int i = 1; i <= steps - 1; i++) {
        victim = victim->next;
    }
    
    while (victim != NULL) {
        Node* temp = victim->next;
        free(victim);
        victim = temp;
        if (victim == NULL) {
            victim = head;
            while (victim->next != head) {
                victim = victim->next;
            }
        } else {
            for (int i = 1; i < steps; i++) {
                victim = victim->next;
            }
        }
    }
}

// 添加节点到链表
void appendToList(Node** head, int value) {
    Node* newNode = createNode(value);
    if (*head == NULL) {
        *head = newNode;
    } else {
        Node* current = *head;
        while (current->next != NULL) {
            current = current->next;
        }
        current->next = newNode;
    }
}

int main() {
    Node* head = NULL;
    int n, steps;
    printf("Enter the number of people: ");
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        appendToList(&head, i);
    }
    printf("Enter the step count: ");
    scanf("%d", &steps);

    josephusProblem(head, steps);
    printf("The last person standing is at position %d\n", head->data);

    return 0;
}

在这个代码里，用户先输入人数和步骤数，然后程序会创建一个循环链表，并应用约瑟夫环算法。`josephusProblem`函数负责执行约瑟夫环的规则，最后输出存活的人及其位置。