UVa 621 Secret Research （water ver.）

UVa 621 Secret Research （water ver.）2014-07-16 csdn博客 synapse7621 - Secret Research

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php？option=com_onlinejudge&Itemid=8&category=99&page=show_problem&problem=562

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 then consecutive 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

478783519078944

Sample Output

+-？*

简单匹配就行。

完整代码：

/*0.013s*/#include<cstdio>#include<cstring>char str[10010];int main（void）{int n, len;scanf（"%d", &n）;while （n--）{scanf（"%s", str）;len = strlen（str）;if （！strcmp（str, "1"） || ！strcmp（str, "4"） || ！strcmp（str, "78"））puts（"+"）;else if （str[len - 2] == "3" && str[len - 1] == "5"）puts（"-"）;else if （str[0] == "9" && str[len - 1] == "4"）puts（"*"）;else //str[0] == "1" && str[1] == "9" && str[2] == "0"puts（"？"）;}return 0;}