621 - Secret Research
| Secret Research |
At a certain laboratory results of secret research are thoroughly encrypted. A result of a single experiment is stored as an information of its completion:
`positive result',`negative result',`experiment failed' or`experiment not completed'
The encrypted result constitutes a string of digits S, which may take one of the following forms:
positive result S = 1 or S = 4 or S = 78
negative result S = S35
experiment failed S = 9S4
experiment not completed S = 190S
(A sample result S35 means that if we add digits 35 from the right hand side to a digit sequence then we shall get the digit sequence corresponding to a failed experiment)
You are to write a program which decrypts given sequences of digits.
Input
A integer
n stating the number of encrypted results and thenconsecutive
n lines, each containing a sequence of digits given as ASCII strings.
Output
For each analysed sequence of digits the following lines should be sent to output (in separate lines):
+ for a positive result
- for a negative result
* for a failed experiment
? for a not completed experiment
In case the analysed string does not determine the experiment result, a first match from the above list should be outputted.
Sample Input
4
78
7835
19078
944
Sample Output
+
-
?
*
#include<stdio.h>
#include<string.h>
int main()
{
int n;
scanf("%d",&n);
while(n--)
{
char a[1005]={0};
scanf("%s",a);
if(strlen(a)<=2)
puts("+");
else if(a[strlen(a)-1]=='5'&&a[strlen(a)-2]=='3')
puts("-");
else if(a[0]=='9'&&a[strlen(a)-1]=='4')
puts("*");
else if(a[0]=='1'&&a[1]=='9'&&a[2]=='0')
puts("?");
}
return 0;
}
621 - Secret Research的更多相关文章
- UVA题目分类
题目 Volume 0. Getting Started 开始10055 - Hashmat the Brave Warrior 10071 - Back to High School Physics ...
- Volume 1. Maths - Misc
113 - Power of Cryptography import java.math.BigInteger; import java.util.Scanner; public class Main ...
- poj3708(公式化简+大数进制装换+线性同余方程组)
刚看到这个题目,有点被吓到,毕竟自己这么弱. 分析了很久,然后发现m,k都可以唯一的用d进制表示.也就是用一个ai,和很多个bi唯一构成. 这点就是解题的关键了. 之后可以发现每次调用函数f(x),相 ...
- poj 3708 Recurrent Function
Recurrent Function Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 1233 Accepted: 336 ...
- PHP serialize && unserialize Security Risk Research
目录 . 序列化的定义 . serialize:序列化 . unserialize:反序列化 . 序列化.反序列化存在的安全风险 . Use After Free Vulnerability -] . ...
- The top 100 papers Nature explores the most-cited research of all time.
The top 100 papers Nature explores the most-cited research of all time. The discovery of high-temper ...
- 【oneday_onepage】——The Secret Of Steve<1>
The Secret Of SteveThe secret of Steve is simple. It explains his success and excess. It exemplifies ...
- Android Secret Code
我们很多人应该都做过这样的操作,打开拨号键盘输入*#*#4636#*#*等字符就会弹出一个界面显示手机相关的一些信息,这个功能在Android中被称为android secret code,除了这些系 ...
- ASP.NET OAuth:access token的加密解密,client secret与refresh token的生成
在 ASP.NET OWIN OAuth(Microsoft.Owin.Security.OAuth)中,access token 的默认加密方法是: 1) System.Security.Crypt ...
随机推荐
- cocos2dx进阶学习之CCSpriteBatchNode
继承关系 CCSpriteBatchNode -> CCNode, CCTextureProtocol 成员变量 inline CCTextureAtlas* getTextureAtlas(v ...
- fedora 下安装 文泉驿正黑体
1. 可以到文泉驿正黑体的 主页 http://wenq.org/wqy2/index.cgi?%E9%A6%96%E9%A1%B5 最好能去官网表示一下支持 2. 也可以直接使用命令 : sudo ...
- VS2010/MFC对话框四:为控件添加消息处理函数
为控件添加消息处理函数 创建对话框类和添加控件变量在上一讲中已经讲过,这一讲的主要内容是如何为控件添加消息处理函数. MFC为对话框和控件等定义了诸多消息,我们对它们操作时会触发消息,这些消息最终由消 ...
- php调用com组件配置 以openoffice为例
什么是com组件? COM构架下,人们可以开发出各种各样的功能专一的组件,然后将它们按照需要组合起来,构成复杂的应用系统.COM与语言,平台无关的特性使所有的程序员均可充分发挥自己的才智与专长编写组件 ...
- CentOS下mysql最大连接数设置 1040 too many connection
当最大连接数比較小时,可能会出现"1040 too many connection"错误. 能够通过改动配置文件来改动最大连接数,但我连配置文件在哪都不知道,应该怎么办呢? 首先须 ...
- C# 中datagridview行里面有三个cheeckbox,要控制成三选一。
我之前有试过在cellendedit中处理,可以达成效果,当不符合用户打单的界面要求.该事件是在单元格编辑结束之后, 当用户选中两个checkbox,且焦点不移开时,界面上会出现有两个checkbox ...
- qq邮箱是怎么做到同一个浏览器让多个不用用户同时打开的? --session的控制
待解:..... 借鉴网址:http://www.zhihu.com/question/20235500 欢迎来讨论.....
- Android中使用"running services"查看service进程内存
从Android 2.0开始,在Settings中加入了一个新的activity("Running Services" activity),它用于显示当前运行的每个Services ...
- 【BZOJ1132】【POI2008】Tro 计算几何 叉积求面积
链接: #include <stdio.h> int main() { puts("转载请注明出处[辗转山河弋流歌 by 空灰冰魂]谢谢"); puts("网 ...
- 分析一个socket通信: server/client
分析一个socket通信: server/client1 server 1. 创建一个server_socket文件,并绑定端口,然后监听端口 (socket, bind, listen) 2. 查询 ...
