[转]用单向循环链表解决约瑟夫环问题

设有n个人围坐一圈,现以某个人开始报数,数到m的人出列,接着从出列的下一个人开始重新报数,数到m的人又出列,如此下去,直到所有人都出列为止.按出列顺序输出.
这段代码是从网上找来的，在此特别说明！！！！！

C代码

#include "stdlib.h"

struct
ele {

int
no;

struct
ele *link;

} main() {

struct
ele *h, *u, *p;

int
n, m, i;

printf("Please input n&m:/n"
);

scanf("%d%d"
, &n, &m);
/*输入n和m*/

h = u = (struct
ele *) malloc(
sizeof
(
struct
ele));
/*形成首表元*/

h->no = 1;

for
(i = 2; i <= n; i++)
/*形成其余的n-1个表元*/

{

u->link = (struct
ele *) malloc(
sizeof
(
struct
ele));

u = u->link;

u->no = i;/*第i个表元置编号i*/

}

u->link = h;/*末表元后继首表元，形成环*/

puts(

"/nThe numbers of who will quit the cycle in turn are:"
);

while
(n) {

for
(i = 1; i < m; i++)

/*掠过m－1个表元*/

u = u->link;

p = u->link;/*p指向第m个表元*/

u->link = p->link;/*第m个表元从环中脱钩*/

printf("%4d"
, p->no);

free(p);/*释放第m个表元占用的空间*/

n--;

}

printf("/n/n Press any key to quit.../n"
);

getchar();

}

#include   "stdlib.h"
struct ele {
int no;
struct ele *link;
} main() {

struct ele *h, *u, *p;
int n, m, i;
printf("Please   input   n&m:/n");
scanf("%d%d", &n, &m);/*输入n和m*/
h = u = (struct ele *) malloc(sizeof(struct ele));/*形成首表元*/
h->no = 1;
for (i = 2; i <= n; i++)/*形成其余的n-1个表元*/
{
u->link = (struct ele *) malloc(sizeof(struct ele));
u = u->link;
u->no = i;/*第i个表元置编号i*/
}
u->link = h;/*末表元后继首表元，形成环*/
puts(
"/nThe   numbers   of   who   will   quit   the   cycle   in   turn   are:");
while (n) {
for (i = 1; i < m; i++)
/*掠过m－1个表元*/
u = u->link;
p = u->link;/*p指向第m个表元*/
u->link = p->link;/*第m个表元从环中脱钩*/
printf("%4d", p->no);
free(p);/*释放第m个表元占用的空间*/
n--;
}
printf("/n/n   Press   any   key   to   quit.../n");
getchar();
}

当碰到问题,首先考虑这个问题是怎样的类型,第二从解决列表中选择中最合适的方案.以这个问题为例:问题中是一个循环性质，关键点“有n个人围坐一圈”。
可选数据结构与算法中的可选方案：队列，栈，线性表，串。队列，栈和串都不大合适。以线性表中的单向循环链表最合适。