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分享C语言双向约瑟夫问题，输出那边不清楚哪里出错了，free也有问题，求各位大神帮忙看一下程序TUT
约瑟夫问题是一个经典的问题，我们不妨将这个经典问题进行扩展，变成一个双向的约瑟夫问题。
已知n个人（不妨分别以编号1，2，3，…，n 代表 ）围坐在一张圆桌周围，首先从编号为 k 的人从1开始顺时针报数，1, 2, 3, ...，记下顺时针数到 m 的那个人，同时从编号为 k 的人开始逆时针报数，1, 2, 3, ...，数到 m 后，两个人同时出列。然后从出列的下一个人又从 1 开始继续进行双向报数，数到m的那两个人同时出列，…；。依此重复下去，直到圆桌周围的人全部出列。直到圆桌周围只剩一个人为止。
如果双向报数报到 m 时落在同一个人身上，那本次出列的只有一个人。
例如：5,1,2。则总共5个人，从1开始顺时针报数，数到2，定位编号2；同时从1开始报数数到2，定位编号5；2和5同时出列。然后继续开始报数，顺时针报数3,4，定位到4；逆时针报数4,3，定位3；4和3同时出列。最后剩余的为编号1。输出为：2-5,4-3,1,。
如果输入：6,2,3。则输出：4-6,2,1-3,5,。其中第2次只输出一个2，表示第二次双向报数时，恰好都落在编号2上，所以只有一个编号出列。
输入
n,k,m
输出
按照出列的顺序依次输出编号。同时出列编号中间用减号“-”连接。
非法输入的对应输出如下
a)
输入：n、k、m任一个为0
输出：n,m,k must bigger than 0.
b)
输入：k>n
输出：k should not bigger than n.
测试用例1：6,2,3 结果：4-6,2,1-3,5,
测试用例2：5,1,2 结果：2-5,4-3,1,
程序
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
struct seat{
int num;
struct seat * prior;
struct seat * next;
};
int main()
{
int k,n,m,i,count=1,j,numl=0,numr=0;
scanf ("%d,%d,%d",&n,&k,&m);
// k是初始座位，m是数的个数
struct seat * head ,* p,* q,* tail,* r,*s;
head = (struct seat * ) malloc (sizeof(struct seat));
tail = (struct seat * ) malloc (sizeof(struct seat));
head->num = -1;
head->next = head;
head->prior = head;
if(k<1||n<1||m<1)
printf("n,m,k must bigger than 0.\n");
else if (k>n)
printf("k should not bigger than n.\n");
else
{
for (i=n;i>0;i--)
{
p = (struct seat *)malloc(sizeof(struct seat));
p->num = i;
p->prior = head;
p->next = head->next;
head->next = p;
p->next->prior = p;
}
while (p->next != head)
p = p->next;
p->next = head->next;
p->next->prior = p;
while (p->num != k)
p = p->next;
r = p;
while (n)
{
for (j=1;j<m;j++)
p = p->next;
q = p;
numr = q->num;
for (j=1;j<m;j++)
r = r->prior;
s = r;
numl = s->num;
if (numr==numl)
{
printf("%d",numr);
n--;
q->prior->next = q->next;
q->next->prior = q->prior;
free(q);
}
else
{
printf("%d-%d",numr,numl);
n=n-2;
q->prior->next = q->next;
q->next->prior = q->prior;
free(q);
s->prior->next = s->next;
s->next->prior = s->prior;
free(s);
}
if (n != 0) printf(",");
}
printf("\n");
}
return 0;
}
已知n个人（不妨分别以编号1，2，3，…，n 代表 ）围坐在一张圆桌周围，首先从编号为 k 的人从1开始顺时针报数，1, 2, 3, ...，记下顺时针数到 m 的那个人，同时从编号为 k 的人开始逆时针报数，1, 2, 3, ...，数到 m 后，两个人同时出列。然后从出列的下一个人又从 1 开始继续进行双向报数，数到m的那两个人同时出列，…；。依此重复下去，直到圆桌周围的人全部出列。直到圆桌周围只剩一个人为止。
如果双向报数报到 m 时落在同一个人身上，那本次出列的只有一个人。
例如：5,1,2。则总共5个人，从1开始顺时针报数，数到2，定位编号2；同时从1开始报数数到2，定位编号5；2和5同时出列。然后继续开始报数，顺时针报数3,4，定位到4；逆时针报数4,3，定位3；4和3同时出列。最后剩余的为编号1。输出为：2-5,4-3,1,。
如果输入：6,2,3。则输出：4-6,2,1-3,5,。其中第2次只输出一个2，表示第二次双向报数时，恰好都落在编号2上，所以只有一个编号出列。
输入
n,k,m
输出
按照出列的顺序依次输出编号。同时出列编号中间用减号“-”连接。
非法输入的对应输出如下
a)
输入：n、k、m任一个为0
输出：n,m,k must bigger than 0.
b)
输入：k>n
输出：k should not bigger than n.
测试用例1：6,2,3 结果：4-6,2,1-3,5,
测试用例2：5,1,2 结果：2-5,4-3,1,
程序
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
struct seat{
int num;
struct seat * prior;
struct seat * next;
};
int main()
{
int k,n,m,i,count=1,j,numl=0,numr=0;
scanf ("%d,%d,%d",&n,&k,&m);
// k是初始座位，m是数的个数
struct seat * head ,* p,* q,* tail,* r,*s;
head = (struct seat * ) malloc (sizeof(struct seat));
tail = (struct seat * ) malloc (sizeof(struct seat));
head->num = -1;
head->next = head;
head->prior = head;
if(k<1||n<1||m<1)
printf("n,m,k must bigger than 0.\n");
else if (k>n)
printf("k should not bigger than n.\n");
else
{
for (i=n;i>0;i--)
{
p = (struct seat *)malloc(sizeof(struct seat));
p->num = i;
p->prior = head;
p->next = head->next;
head->next = p;
p->next->prior = p;
}
while (p->next != head)
p = p->next;
p->next = head->next;
p->next->prior = p;
while (p->num != k)
p = p->next;
r = p;
while (n)
{
for (j=1;j<m;j++)
p = p->next;
q = p;
numr = q->num;
for (j=1;j<m;j++)
r = r->prior;
s = r;
numl = s->num;
if (numr==numl)
{
printf("%d",numr);
n--;
q->prior->next = q->next;
q->next->prior = q->prior;
free(q);
}
else
{
printf("%d-%d",numr,numl);
n=n-2;
q->prior->next = q->next;
q->next->prior = q->prior;
free(q);
s->prior->next = s->next;
s->next->prior = s->prior;
free(s);
}
if (n != 0) printf(",");
}
printf("\n");
}
return 0;
}
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nihuayewoa 2017-10-20
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好的谢谢~
自信男孩 2017-10-18
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#include <stdio.h>
#include <string.h>
#include <stdlib.h>
struct seat{
int num;
struct seat * prior;
struct seat * next;
};
void create_link(struct seat **pphead, int n)
{
int i;
struct seat *p, *pcur;
if (!(*pphead)) {
*pphead = (struct seat * ) malloc (sizeof(struct seat));
if (!(*pphead)) {
fprintf(stderr, "malloc error!\n");
exit(0);
}
}
(*pphead)->num = n;
(*pphead)->next = *pphead;
(*pphead)->prior = *pphead;
pcur = *pphead;
for (i = n-1; i > 0; i--) {
p = (struct seat *)malloc(sizeof(struct seat));
if (!p) {
fprintf(stderr, "malloc error!\n");
exit(0);
}
p->num = i;
p->prior = pcur;
p->next = pcur->next;
pcur->next = p;
pcur = p;
}
}
void show_list(struct seat *phead)
{
struct seat *pcur = phead;
printf("Positive: ");
while (pcur->next != phead) {
printf("%d\t", pcur->num);
pcur = pcur->next;
}
printf("%d\n", pcur->num);
printf("Reverse: ");
while (pcur != phead) {
printf("%d\t", pcur->num);
pcur = pcur->prior;
}
printf("%d\n", pcur->num);
}
struct seat *delete_link(struct seat *phead, int start, int num)
{
struct seat *pcur = phead, *pdel;
int i = 0;
while (i++ < start)
pcur = pcur->next;
i = 0;
while (pcur->next != pcur) {
if ((i+1) % num == 0) {
pdel = pcur;
pcur->prior->next = pcur->next;
pcur->next->prior = pcur->prior;
printf("%d is out!\n", pdel->num);
pcur = pcur->next;
i++;
free(pdel);
} else {
i++;
pcur = pcur->next;
}
}
return pcur;
}
int main(void)
{
int n, start, m;
struct seat *phead = NULL, *last = NULL;
printf("Please input n, start and m(n,start,m): ");
scanf("%d,%d,%d", &n, &start, &m);
if (start > n) {
fprintf(stdout, "num shold be bigger than start!\n");
return 0;
}
create_link(&phead, n);
show_list(phead);
last = delete_link(phead, start, m);
printf("Last node: %d\n", last->num);
free(last);
phead = NULL;
return 0;
}
