ARITH2  Simple Arithmetics II
While browsing aimlessly, Peter stumbled upon an old riddle he used to solve on his calculator when he was still young. It was the kind of a riddle where you punch in a bunch of numbers and operators into a simple pocket calculator and then turn it upside down to get the answer:
These come in many different sizes but they are always exactly one foot long. Answer: 103 * 103 * 5.
What are made of ice to keep people warm? Answer: 50 * 40 * 250 + 791.
After a few minutes he found a large amount of such riddles and full of excitement he went to solve them. He turned his computer screen upside down...
... only to find out that he does not have a reasonable calculator program installed on his computer.
Problem specification
You are given multiple sequences of button presses of a simple pocket calculator that consist of digits and arithmetic operators. For each such sequence find the number it would produce on a pocket calculator's display.
Note that the pocket calculator evaluates the operators in the order in which they are given. (i.e., there is no operator precedence.) Assume that the display of the calculator is large enough to show the result, and that its memory is sufficient to store all intermediate results.
Input specification
The first line of the input file contains an integer T specifying the number of test cases. Each test case is preceded by a blank line.
Each test case represents one sequence of button presses for a pocket calculator. The sequence consists of nonnegative integers and arithmetic operators and ends with an equal sign. It may also contain spaces to improve readability.
The operator / represents integer division, rounded down. You may assume that no test case contains division by zero and that in all test cases all intermediate results are nonnegative.
Tip: long long int in C/C++, long in Java or int64 in Pascal is enough for this problem.
Output specification
For each sequence from the input file output the number that would be displayed on the calculator.
Example
Input: 4 1 + 1 * 2 = 29 / 5 = 103 * 103 * 5 = 50 * 40 * 250 + 791 = Output: 4 5 53045 500791
Hint
The first test case shows that there is no operator precedence.
The second one shows that integer division always rounds down.
The last two outputs are the answers to the two riddles in the problem statement: "shoes" (53045 upside down), and "igloos"(500791 upside down).
hide comments
ajay_5097:
20160517 15:48:27
Simple implementation!!! Do whatever question is asking nothing extra.


mohitgupta07:
20160514 22:53:52
A Big LOL to those who think using strings and java will make it complicated..i think they need to learn more abt it!


lalit_nit:
20160418 07:05:03
Lol... what kind of Difficulty in java???....@junaid 

marshmellow:
20160405 07:57:17
AC in one go... :D


abc_c:
20160103 15:38:50
got AC in 2nd attempt


Junaid:
20151223 23:30:48
Green light on the first go:)...too easy to solve in C++...JAVA guys can face difficulty here.... 

Timmy Jose:
20151212 14:15:36
Nasty problem  deceptively simple. Taught me a lot about handling extreme cases of string input though, so it's all good! 

chandan kumar:
20151125 08:46:26
Using strings may make the code complicated. just scan numbers directly and inbetween numbers, scan the operators. 

avidcoder:
20151109 12:48:56
good problem. Don't use strings. 

sumit suthar:
20151104 10:30:29
Easy one :)) 
Added by:  Fudan University Problem Setters 
Date:  20090531 
Time limit:  2s 
Source limit:  50000B 
Memory limit:  1536MB 
Cluster:  Cube (Intel G860) 
Languages:  All except: C99 ERL JSRHINO NODEJS PERL6 
Resource:  IPSC 2009 