约瑟夫环 出列顺序中遇到的free（）函数问题
#include<stdio.h>
#include<stdlib.h>
typedef struct node{
int a;
int b;
struct node *next;
}Lnode;
Lnode *creatlist(int n);
void printlist(Lnode *p);
void fun(Lnode *p,int n,int m);
main()
{ Lnode *p;
int n,m;
printf("input n and m:\n");
scanf("%d,%d",&n,&m);
p=creatlist(n);
printf("the list is:\n");
printlist(p);
printf("\nchu lie sunxu:\n");
fun(p,n,m);
}
Lnode *creatlist(int n)
{ int i;
Lnode *head,*p,*s;
head=( Lnode *) malloc(sizeof(Lnode ));
p=head;
head->next=p;
printf("enter a,b:\n");
scanf ("%d,%d",&p->a,&p->b);
for (i=2; i<=n; ++i )
{ s=(Lnode *) malloc(sizeof(Lnode ));
scanf ("%d,%d",&s->a,&s->b);
s->next=head;
p->next=s; p=s;
}
return (s);
}
void printlist(Lnode *p)
{ Lnode *q;
q=p->next;
while(q!=p)
{ printf("%3d",q->b);
q=q->next;
}
printf("%3d",p->b);
}
void fun(Lnode *p,int n,int m)
{ Lnode *q;
int i,count;
q=p;
for(i=1;i<=n;i++)
{ count=0;
while(count!=m)
{ p=q;
q=q->next;
count++;
}
printf("%3d",q->a);
m=q->b;
p->next=q->next; free(q);
}
}

[解决办法]

C/C++ code

//假设有n个人团团围做，从第1个人开始数数，数到第m个人时候，第m个人出列，//然后继续从1开始数数，数到第m个人退出#include <stdio.h>#include <conio.h>int i,k,t;int n,m;static char f[1001];//0该座位未出圈,1该座位已出圈void main() {    while (1) {        printf("Input n m(1000>=n>=m>=1):");        fflush(stdout);        rewind(stdin);        if (2==scanf("%d%d",&n,&m)) {            if (1000>=n && n>=m && m>=1) break;        }    }    t=0;//已出圈总人数    i=1;//座位编号    k=1;//当前要数的数    while (1) {        if (0==f[i]) {            if (m==k) {                t++;                f[i]=1;                printf("%3d ",i);                if (0==t%10) printf("\n");                if (t>=n) break;            }            k++;if (k>m) k=1;        }        i++;if (i>n) i=1;    }    cprintf("Press any key ...");    getch();}
[解决办法]
#include"stdio.h"
#include<stdlib.h>
typedef struct node{
 int b;//结点位序  
 struct node *next;
 }Lnode,LinkList;

LinkList *creatlist(int n)
{  
 int i;
 LinkList *head,*s,*tail;
 head=( Lnode *)malloc(sizeof(Lnode));
 head = NULL;
 for (i=1;i<=n;i++)
 {   
 s=(LinkList *)malloc(sizeof(Lnode ));
 if (s == NULL)
 {
printf("申请失败\n"); 
}

 s->b = i;
 if (head == NULL)
 {
 head = s;
 tail = head;
 } 
 else  
 {
   tail->next = s;
   tail = s;
}
   tail->next = head;  
  
  }
  return head;
}

void fun(LinkList *head,int n,int m)
{  
   LinkList *q,*pre;
   int count=0;
   int num = 0;//记录已经出列的人数  
   q=head;
while(num != n)
{  
if (count!=m)
{
pre = q;
q=q->next;
count++;
}
count = 0;
printf("%3d",q->b);   
 
pre->next = q->next;
num++;
free(q);
q = pre->next;
}
}  


int main()
{  
 LinkList *head;
 int n,m;
 printf("input n and m:\n");
 scanf("%d%d",&n,&m);
 head=creatlist(n);//建立一个带头结点的链表  
 printf("\nchu lie sunxu:\n");
 fun(head,n,m);
 return 0;
}

改了点
[解决办法]
我也做了一个你看看
C/C++ code
#include <stdio.h>#include <stdlib.h>/*约瑟夫环是一个数学的应用问题：已知n个人（以编号1，2，3...n分别表示）围坐在一张圆桌周围。从编号为k的人开始报数，数到m的那个人出列；他的下一个人又从1开始报数，数到m的那个人又出列；依此规律重复下去，直到圆桌周围的人全部出列。n=9,k=1,m=5【解答】出局人的顺序为5, 1, 7, 4, 3, 6, 9, 2, 8。*/struct list{    int n;    struct list *netx;};typedef struct list List;int peo = 9,number = 1,out=5;void display(List *p);void outdisy(List *p);int main(){    int i;    int counter;//技术器    List *head =NULL,*current,*p;    for(i = 1;i <= peo;i++)    {        current =(List *)malloc(sizeof(List));        if(head == NULL)        {            head = current;            current->n = i;        }        else        {            current->n = i;            if(i == peo)                current->netx = head;            p->netx = current;        }        p = current;    }    printf("head地址:%x\n",head);//    display(head);    outdisy(head);//出列显示    printf("Hello world!\n");    return 0;}void display(List *p){    List *temp = p;    do    {        printf("%d\t%x\n",temp->n,temp->netx);        temp = temp->netx;    }while(temp != p);}void outdisy(List *p){    int i;    int count = 0,suni=1;    List temp,*current;    printf("out:\n");    while(count!= peo)    {        if(suni == 4)            current = p;        if(suni == out)        {            printf("%d\t%x\n",p->n,p->netx);            temp = *p;            free(p);            p=temp.netx;            current->netx=p;            suni = 1;            count++;        }        suni++;        p = p->netx;    }}