博客
关于我
强烈建议你试试无所不能的chatGPT,快点击我
Poj1218--THE DRUNK JAILER
阅读量:6148 次
发布时间:2019-06-21

本文共 2337 字,大约阅读时间需要 7 分钟。

 

THE DRUNK JAILER
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 25123   Accepted: 15766

Description

A certain prison contains a long hall of n cells, each right next to each other. Each cell has a prisoner in it, and each cell is locked.
One night, the jailer gets bored and decides to play a game. For round 1 of the game, he takes a drink of whiskey,and then runs down the hall unlocking each cell. For round 2, he takes a drink of whiskey, and then runs down the
hall locking every other cell (cells 2, 4, 6, ?). For round 3, he takes a drink of whiskey, and then runs down the hall. He visits every third cell (cells 3, 6, 9, ?). If the cell is locked, he unlocks it; if it is unlocked, he locks it. He
repeats this for n rounds, takes a final drink, and passes out.
Some number of prisoners, possibly zero, realizes that their cells are unlocked and the jailer is incapacitated. They immediately escape.
Given the number of cells, determine how many prisoners escape jail.

Input

The first line of input contains a single positive integer. This is the number of lines that follow. Each of the following lines contains a single integer between 5 and 100, inclusive, which is the number of cells n.

Output

For each line, you must print out the number of prisoners that escape when the prison has n cells.

Sample Input

25100

Sample Output

210

Source

当年的开关灯问题变成了“高富帅”, 差点没读懂 。 RT.
1 #include 
2 #include
3 #include
4 using namespace std; 5 int num[110]; 6 int n; 7 int main() 8 { 9 int t;10 scanf("%d", &t);11 while(t--)12 {13 scanf("%d", &n);14 for(int i = 1; i <= n; i++) 15 num[i] = 1;16 for(int i = 1; i <= n; i++)17 { 18 for(int j = i; j <= n; j+=i)19 {20 if(num[j]==0)21 {22 num[j] = 1;23 continue;24 }25 if(num[j]==1)26 {27 num[j] = 0;28 continue;29 }30 }31 } 32 int sum = 0;33 for(int i = 1; i <= n; i++)34 if(!num[i])35 sum++;36 printf("%d\n", sum);37 } 38 }

 

 

转载于:https://www.cnblogs.com/soTired/p/4742556.html

你可能感兴趣的文章
011-Spring Boot 运行流程分析SpringApplication.run
查看>>
[解决方案]sql server复制需要有实际的服务器名称才能连接到服务器
查看>>
用yarn替代npm
查看>>
ReSharper2017.3的列对齐、排版格式、列对齐错误的修复
查看>>
BZOJ3273 : liars
查看>>
mysql开发之---每日一得01
查看>>
Java基础-MySQL数据库扫盲篇
查看>>
Echarts关于仪表盘
查看>>
php实现求数组中出现次数超过一半的数字(isset($arr[$val]))(取不同数看剩)(排序取中)...
查看>>
(转)创业的注意事项
查看>>
超简单的视频对象提取程序
查看>>
『TensorFlow』线程控制器类&变量作用域
查看>>
【知识小结】Git 个人学习笔记及心得
查看>>
已超过传入消息(65536)的最大消息大小配额。若要增加配额,请使用相应绑定元素上的 MaxReceivedMessageSize 属性...
查看>>
『流畅的Python』第9章笔记_对象
查看>>
结构化数据、半结构化数据和非结构化数据
查看>>
SQLServer 2014 本地机房HA+灾备机房DR解决方案
查看>>
centos7安装配置mysql5.6
查看>>
反爬虫:利用ASP.NET MVC的Filter和缓存(入坑出坑)
查看>>
SaaS技术栈的走势
查看>>