#include<stdio.h>
#include<stdlib.h>
struct node{
int num;
struct node *next;
}a;
struct node * creat(int n); //返回头指针
int excute(int m,struct node *head);//返回最后剩下的人的编号
main()
{
int n;
int m;
struct node *h = NULL;
printf("input how many people\n");
scanf("%d",&n);
printf("input which people been delete\n");
scanf("%d",&m);
if(n < 1 || m < 1){ /*对特殊情况的讨论*/
printf("wrong input!\n");
return -1;
}else if( n == 1 ){
printf("nobady left!\n");
return 0;
}else if( m == 1 ){
printf("the last number is %d\n",n);
return 0;
}else{
h = creat(n);
printf("\nthe last number is %d\n",excute(m,h));
return 0;
}
}
struct node * creat(int n) //构造循环链表
{
struct node * p = NULL;
struct node * h = NULL;
struct node * q = NULL;
int num = 0;
p = (struct node *)malloc( sizeof(a) );
p->num = ++num;
p->next = NULL;
h = p;
while( --n !=0 ){
q= (struct node *)malloc( sizeof(a) );
q->num = ++num;
p->next = q;
p = q;
}
if(p != NULL)
p->next = h;
return h;
}
int excute(int m,struct node *h)
{
struct node *current=NULL;
struct node * p=NULL;
int i;
current = h;
printf("the out order is:");
while (current->next != current){
for (i = 1; i < m; i++){ /* 查找报数为m的小孩结点 */
p = current;
current = current->next;
}
p->next = current->next; /* 删除报数为m的小孩结点 */
printf("%5d",current->num); //打印出被淘汰的顺序
free(current); /* 释放报数为m的小孩结点空间 */
current = p->next;
}
return current->num;
}
1，先确定数据结构为循环链表，每一个元素为一个结构，包括两个变量。再画一个图，找到解题的思路，写出大致的算法。
2，根据画的图，先不考虑特殊的情况，写出代码。
3，编译，链接，确保无错。
4，加入安全方面的考虑，如非法输入等。
5，加入易用性方面的代码。
6，编译，链接，确保无错。
之后可以参考一下别人的代码，可能会发现很多新的方法和思路。
比如，这个题目也可以这样做：
#include <stdio.h>
#include <stdlib.h>
#include <string.h> //没有使用链表
int main(int argc, char *argv[])
{
int m, n, step=0, index=0, cir=0;
char *out=NULL;
scanf("%d%d", &m, &n);
out = (char*)malloc(m);//可以看作是数组的动态分配，好！！。与memset函数联合使用
if(out == NULL)
return -1;
memset(out, 0, m); //因为malloc不对分配的内存进行初始化，所以这里用memset进行初始化
while(1){
if(step==m)
break;
while(cir<n) {
if(!out[index])
++cir;
index=(index+1)%m;
}
cir=0;
++step;
out[(index+m-1)%m]=1;
printf("\nNo.%d out.", !index?m:index);
}
free(out);
system("PAUSE"); //当有非法输入时，如0，5或4，0时，在屏幕上显示“请按任意键继续”
return 0;
}
也可以这样做：
/**********************************************
********* Josephus problem ************
*********************************************/ //和我的方法差不多
#include <malloc.h>
#include <stdio.h>
struct boy /* struct of boy*/
{
int num;
struct boy *next;
};
int main()
{
int i;
int number; /* 小孩的数目 */
int interval; /* 即报数从1- interval */
struct boy *head, *p, *current;
printf("please input the number of boys:");
scanf("%d", &number);
printf("please input the interval:");
scanf("%d", &interval);
if(number < interval || number <= 0 || interval <= 0)
{
printf("input error!!!\n");
return 0;
}
head = (struct boy *) malloc(sizeof(struct boy));
head->num = 1;
head->next = NULL;
current = head;
printf("Boy numbers:%d,", head->num);
for (i = 2; i <= number; i++)/* 建立单向循环链表 */
{
p = (struct boy *) malloc(sizeof(struct boy));
p->num = i;
p->next = NULL;
current->next = p;
current = p;
printf(" %d,", p->num);
}
printf("\n");
current->next = head;
current = head;
printf("Out queue order:");
while (current->next != current) /*loop to find the winner */
{
for (i = 1; i < interval; i++) /* 查找报数为interval的小孩结点 */
{
p = current;
current = current->next;
}
p->next = current->next; /* 删除报数为interval的小孩结点 */
printf(" %d,", current->num);
free(current); /* 释放报数为interval的小孩结点空间 */
current = p->next;
}
printf("\n");
printf("No.%d boy's won!\n", current->num);
getch();
return 0;
}
