C语言
约瑟夫环算法
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我也要 关注 + 加入社区类似于以下问题可以用约瑟夫环的算法来解决：
例：用户输入M,N值，从1至N开始顺序循环数数，每数到M输出该数值(这个值在以后的循环中不再数到)，直至说有数据全部输出。
#include <stdio.h>
#include <malloc.h>
typedef struct Node
{
int index;
struct Node *next;
}JosephuNode;
int Josephu(int n, int m)
{
int i, j;
JosephuNode *head, *tail;
head = tail = (JosephuNode *)malloc(sizeof(JosephuNode));
for (i = 1; i < n; ++i)
{
tail->index = i;
tail->next = (JosephuNode *)malloc(sizeof(JosephuNode));
tail = tail->next;
}
tail->index = i;
tail->next = head;
for (i = 1; tail != head; ++i)
{
for (j = 1; j < m; ++j)
{
tail = head;
head = head->next;
}
tail->next = head->next;
printf("第%4d个出局的人是：%4d号\n", i, head->index);
free(head);
head = tail->next;
}
i = head->index;
free(head);
return i;
}
int main()
{
int n, m;
scanf("%d%d", &n, &m);
printf("最后胜利的是%d号！\n", Josephu(n, m));
//system("pause");
return 0;
}
例：用户输入M,N值，从1至N开始顺序循环数数，每数到M输出该数值(这个值在以后的循环中不再数到)，直至说有数据全部输出。
#include <stdio.h>
#include <malloc.h>
typedef struct Node
{
int index;
struct Node *next;
}JosephuNode;
int Josephu(int n, int m)
{
int i, j;
JosephuNode *head, *tail;
head = tail = (JosephuNode *)malloc(sizeof(JosephuNode));
for (i = 1; i < n; ++i)
{
tail->index = i;
tail->next = (JosephuNode *)malloc(sizeof(JosephuNode));
tail = tail->next;
}
tail->index = i;
tail->next = head;
for (i = 1; tail != head; ++i)
{
for (j = 1; j < m; ++j)
{
tail = head;
head = head->next;
}
tail->next = head->next;
printf("第%4d个出局的人是：%4d号\n", i, head->index);
free(head);
head = tail->next;
}
i = head->index;
free(head);
return i;
}
int main()
{
int n, m;
scanf("%d%d", &n, &m);
printf("最后胜利的是%d号！\n", Josephu(n, m));
//system("pause");
return 0;
}
...全文
