两种方法解决约瑟夫问题

第一种：构建循环链表，然后不断遍历链表直到剩下最后一个元素。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<cstdlib>
#include<algorithm>
 
using namespace std;
typedef struct list
 {
     int data;
     struct list *next;
 }list,*link;
  int main(){
    link p = NULL,q = NULL,head = NULL;
     int i,n,m;
     scanf("%d %d",&n,&m);
     head = (link)malloc(sizeof(list));
     q = head;
     for(i = 1;i <= n;i++)        //构造循环链表
     {    
         p = (link)malloc(sizeof(list));
         p -> data = i;
         q -> next = p;
         q = p;
     }    p -> next = head -> next;
     p = head -> next;
     free(head);
     while(p -> next != p)         i < m - 1 是因为第一次循环 p 位于 第一个节点 与第m个节点中间隔有m-2个节点 所以执行m-2次循环
     {
         for(i = 1;i < m - 1;i++)
         {
             p = p -> next;
         }
         p -> next = p -> next -> next;
         p = p -> next;
     }
     printf("%d\n",p -> data);return 0;
}

 

第二种：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<cstdlib>
#include<algorithm>
 
using namespace std;
 
int main(){
    int n,m;
     int p,q,t,i;
     int a[300];
     scanf("%d %d",&n,&m);
     memset(a,0,sizeof(a));
     q = n;
     p = 0;
     for(i = 0;i < n;i++) a[i] = i + 1;
     for(i = 0;;i++)
     {
         if(i == n) i = 0;
         if(a[i] != 0) p++;
         else continue;
         if(p % m == 0)
         {
             a[i] = 0;
             q--;
         }
         if(q == 1) break;
     }
     for(i = 0;i < n;i++)
     {
         if(a[i]) printf("%d\n",a[i]);
     }
  return 0;
}