Solution of 621 - Secret Research

Problem Description
source:https://uva.onlinejudge.org/external/6/621.html

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 

4 
78 
7835 
19078 
944 

Sample Output 

+ 
- 
? 
*

Solution:

#include <algorithm>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <vector>
#include<stdio.h>
#include<stdlib.h>

using namespace std;

int main()
{
    int len,t,i;
    string s;
    while(scanf("%d",&t)==1)
    {
        for(i=1;i<=t;i++)
        {
            cin>>s;
            len=s.size();
            if(s=="1"||s=="4"||s=="78")
                printf("+\n");
            else if(s[len-1]=='5' && s[len-2]=='3')
                printf("-\n");
            else if(s[0]=='9' && s[len-1]=='4')
                printf("*\n");
            else
                printf("?\n");
        }
    }
    return 0;
}