第四周 项目4 猴子选大王

目标：一群猴子，编号是1，2，3，，，，m，这群猴子按照1-m排序做一圈。从第一只开始数，没到第n个，该猴子离开圈子，到最后，剩下一个猴子就是猴王。

代码：

#include <iostream>#include "kudetouwenjian.h"using namespace std;int main(){     linknode *L;     A e=0;     int m,n,c;     cin>>m>>n;     A a[m];     A b[m];     for(int i=0;i<m;i++)          a[i]=i+1;     xunhuanlianbiao(L,a,m);     if(n==1)     {          cout<<m<<endl;          return 0;     }     houzi(L,n,m);     return 0;}#include <iostream>#include <malloc.h>#include "kudetouwenjian.h"using namespace std;void houzi(linknode *&L,int n,int m){     linknode *p=L->next;     for(int i=0;i<m-1;i++)     {          for(int a=0;a<n-1;a++)          {               L=p;               p=L->next;          }          L->next=p->next;          free(p);          p=L->next;     }     cout<<L->shuju<<endl;}void xunhuanlianbiao(linknode *&L,A a[],int n){         linknode *s,*r;     L=(linknode *)malloc(sizeof(linknode));     r=L;     for(int i=0;i<n;i++)     {          s=(linknode *)malloc(sizeof(linknode));          s->shuju=a[i];          r->next=s;          r=s;     }     r->next=L->next;}void toucha(linknode *&L,A a[],int n){     linknode *s;     L=(linknode *)malloc(sizeof(linknode));     L->next=NULL;     for(int i=0;i<n;i++)     {          s=(linknode *)malloc(sizeof(linknode));          s->shuju=a[i];          s->next=L->next;          L->next=s;     }}void shuchu(linknode *L){     linknode *p=L->next;     while(p!=NULL)     {          cout<<p->shuju<<" ";          p=p->next;     }     cout<<endl;}void weicha(linknode *&L,A a[],int n){     linknode *s,*r;     L=(linknode *)malloc(sizeof(linknode));     r=L;     for(int i=0;i<n;i++)     {          s=(linknode *)malloc(sizeof(linknode));          s->shuju=a[i];          r->next=s;          r=s;     }     r->next=NULL;}void chushihua(linknode *&L){     L=(linknode *)malloc(sizeof(linknode));     L->next=NULL;}void cuihui(linknode *&L){     linknode * pre=L,* p=L->next;     while(p!=NULL)     {          free(pre);          pre=p;          p=pre->next;     }     free(pre);     L->next=NULL;}int panduanshifouweikong(linknode *L){     return(L->next==NULL);}int changdu(linknode *L){     int e=0;     linknode *p=L->next;     while(p!=NULL)     {          e++;          p=p->next;     }     return(e);}int shunxuchazhao(linknode * L,int i,A &e){     int j=1;     linknode *p=L->next;     if(i<=0)j=j;;     while(j<i)     {          j++;          p=p->next;     }     if(p==NULL)          j=j;     else     {          e=p->shuju;          return(e);     }}int chazhaodange(linknode *L,A e){     int j=1;     linknode *p=L->next;     if(p==NULL)          return 0;     while(p!=NULL&&e!=p->shuju)     {          p=p->next;          j++;     }     if(p==NULL)          return 0;     else     return(1);}bool charudange(linknode *&L,int i,A e){     int j=0;     linknode * p=L,* s;     if(i<=0)return false;     while(j<i-1&&p!=NULL)     {          j++;          p=p->next;     }     if(p==NULL)          return false;     else     {          s=(linknode *)malloc(sizeof(linknode));          s->shuju=e;          s->next=p->next;          p->next=s;     }}#ifndef KUDETOUWENJIAN_H_INCLUDED#define KUDETOUWENJIAN_H_INCLUDED#include <iostream>typedef int A;typedef struct lianbiao{     A shuju;     struct lianbiao *next;}linknode;void toucha(linknode *&L,A a[],int n);void shuchu(linknode *L);void weicha(linknode *&L,A a[],int n);void chushihua(linknode *&L);void cuihui(linknode *&L);int panduanshifouweikong(linknode *L);int changdu(linknode *L);int shunxuchazhao(linknode *L,int i,A &e);int chazhaodange(linknode *L,A e);bool charudange(linknode *&L,int i,A e);void xunhuanlianbiao(linknode *&L,A a[],int n);void houzi(linknode *&L,int n,int m);#endif // KUDETOUWENJIAN_H_INCLUDED

运行图：

心得：

这次算是一次循环链表的应用题。我把它想的太难了，用链表的删除就行了。