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

  1. UVA题目分类

    题目 Volume 0. Getting Started 开始10055 - Hashmat the Brave Warrior 10071 - Back to High School Physics ...

  2. Volume 1. Maths - Misc

    113 - Power of Cryptography import java.math.BigInteger; import java.util.Scanner; public class Main ...

  3. 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 ...

  4. poj 3708 Recurrent Function

    Recurrent Function Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 1233   Accepted: 336 ...

  5. 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 -] . ...

  6. 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 ...

  7. 【oneday_onepage】——The Secret Of Steve<1>

    The Secret Of SteveThe secret of Steve is simple. It explains his success and excess. It exemplifies ...

  8. 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 ...

  9. 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

  1. 自己写的一个简单的Tab类

    //------------- PS_DOM 功能函数 start----------------var PS_DOM ={ indexOf: function(arr, e){ for(var i= ...

  2. android 破解九宫格

    将目录切换到D:/adb目录下,命令如下 敲入命令 adb shell 然后回车,可以见到如下结果 再敲入命令cd /data/system然后回车, 再执行 rm gesture.key 回车,搞定 ...

  3. BZOJ 1631: [Usaco2007 Feb]Cow Party

    题目 1631: [Usaco2007 Feb]Cow Party Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 491  Solved: 362[Sub ...

  4. Flow Problem(最大流)

    Flow Problem Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others) Tot ...

  5. 2013 ACM/ICPC Asia Regional Chengdu Online---1003

    哈哈哈 #include <iostream> #include <cstring> #include <string> #include <cstdio&g ...

  6. c++11 stl atomic_flag 样例

    Author:DriverMonkey Mail:bookworepeng@Hotmail.com Phone:13410905075 QQ:196568501 測试环境:Win7 64 bit 编译 ...

  7. linux常用命令系列—cp 复制文件与文件夹

    原文地址:http://blog.chinaunix.net/xmlrpc.php?r=blog/article&uid=2272&id=37363 指令名称:cp(copy)功能介绍 ...

  8. CentOS下安装两个或多个Tomcat7

    链接地址:http://lcbk.net/tomcat/1407.html 首先安装JDK 安装之前检查下是否已经安装了openJDK,如果已安装,建议用yum remove 卸载掉. [root@b ...

  9. goahead cgi 及出现的问题解决

    1. route.txt    配置cgi路径 route uri=/cgi-bin dir=/web handler=cgi 2.交叉编译生成cgi goahead  源码路径下  ./test/c ...

  10. iOS 8 强制横屏

    最近用到视频播放功能:(Vitamio, 注:在Build Setting 里面的 Other Link Flag 添加-all_load) iOS 8的屏幕旋转比较坑, 使用以下代码可以强制旋转 - ...