HDU1443 模拟（难）

最新推荐文章于 2021-05-06 00:26:59 发布

转载最新推荐文章于 2021-05-06 00:26:59 发布 · 81 阅读

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原文链接：http://www.cnblogs.com/--ZHIYUAN/p/5951045.html

文章标签：

#java #数据结构与算法

本文探讨了一个基于约瑟夫问题的变形题目，即在2*n个人中，前n个为好人，后n个为坏人的情况下，寻找一个特定的数k，使得按照每k个数移除一个的规则，所有坏人被移除而至少留下一个好人。文章通过代码实现了这一问题的求解。

Joseph

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2453 Accepted Submission(s): 1476

Problem Description

The Joseph's problem is notoriously known. For those who are not familiar with the original problem: from among n people, numbered 1, 2, . . ., n, standing in circle every mth is going to be executed and only the life of the last remaining person will be saved. Joseph was smart enough to choose the position of the last remaining person, thus saving his life to give us the message about the incident. For example when n = 6 and m = 5 then the people will be executed in the order 5, 4, 6, 2, 3 and 1 will be saved.

Suppose that there are k good guys and k bad guys. In the circle the first k are good guys and the last k bad guys. You have to determine such minimal m that all the bad guys will be executed before the first good guy.

Input

The input file consists of separate lines containing k. The last line in the input file contains 0. You can suppose that 0 < k < 14.

Output

The output file will consist of separate lines containing m corresponding to k in the input file.

Sample Input

3

4

0

Sample Output

5

30

Source

ACM暑期集训队练习赛（5）

题意：

给出2*n个数串成环，找出一个k使得每隔k个数就去掉一个数，要去掉后n个数的情况下的最小的k。

代码：

  1 /*
  2 用数组和链表写的两个，费了一晚上时间一直超时。数据只有13个最后只能先跑出来数据再打表提交15S水过去。然而怎么就忘了把这两步合在一起呢？
  3 真是糊涂了。
  4 */
  5 #include<iostream>
  6 #include<cstdio>
  7 using namespace std;
  8 int main()
  9 {
 10     int linklt_point[30];
 11     int ans[15];
 12     int t;
 13     for(int n=1;n<14;n++)
 14     {
 15         if(n==0) break;
 16         for(int i=n+1;;i++)
 17         {
 18             if(i%(2*n)!=0&&i%(2*n)<=n)
 19             continue;
 20             for(int j=1;j<2*n;j++)
 21             {
 22                 linklt_point[j]=j+1;
 23             }
 24             linklt_point[2*n]=1;
 25             int sta=1,pre=2*n;
 26             int sum=0;
 27             int m=2*n;
 28             while(1)
 29             {
 30                 int k=i%m;
 31                 if(k==0) k=m;
 32                 for(int j=1;j<k;j++)
 33                 {
 34                     pre=sta;
 35                     sta=linklt_point[sta];
 36                 }
 37                 if(sta<=n)
 38                 break;
 39                 linklt_point[pre]=linklt_point[sta];
 40                 sta=linklt_point[sta];
 41                 sum++;
 42                 m--;
 43                 if(sum==n)
 44                 break;
 45             }
 46             if(sum==n)
 47             {
 48                 ans[n]=i;
 49                 break;
 50             }
 51         }
 52     }
 53     while(scanf("%d",&t)&&t!=0)
 54     {
 55         printf("%d\n",ans[t]);
 56     }
 57     return 0;
 58 }
 59 
 60 
 61 #include<iostream>
 62 #include<cstdio>
 63 using namespace std;
 64 int main()
 65 {
 66     int a[30];
 67     int n;
 68     int ans[15];
 69     for(int t=1;t<14;t++)
 70     {
 71         for(int i=t+1;;i++)
 72         {
 73             for(int j=1;j<=2*t;j++)
 74             a[j]=1;
 75             int sta=0;
 76             int sum=0;
 77             int m=2*t;
 78             while(1)
 79             {
 80                 int k=i%m;
 81                 if(k==0) k=m;
 82                 int l=0;
 83                 while(l!=k)
 84                 {
 85                     sta++;
 86                     if(sta==2*t+1)
 87                     sta=1;
 88                     if(a[sta]==1)
 89                     l++;
 90                 }
 91                 if(sta<=t)
 92                 break;
 93                 m--;
 94                 a[sta]=0;
 95                 sum++;
 96                 if(sum==t)
 97                 break;
 98             }
 99             if(sum==t)
100             {
101                 ans[t]=i;
102                 break;
103             }
104         }
105     }
106     while(scanf("%d",&n)&&n!=0)
107     {
108         printf("%d\n",ans[n]);
109     }
110     return 0;
111 }

转载于:https://www.cnblogs.com/--ZHIYUAN/p/5951045.html