ADDREV - Adding Reversed Numbers

The Antique Comedians of Malidinesia prefer comedies to tragedies. Unfortunately, most of the ancient plays are tragedies. Therefore the dramatic advisor of ACM has decided to transfigure some tragedies into comedies. Obviously, this work is very hard because the basic sense of the play must be kept intact, although all the things change to their opposites. For example the numbers: if any number appears in the tragedy, it must be converted to its reversed form before being accepted into the comedy play.

Reversed number is a number written in Arabic numerals but the order of digits is reversed. The first digit becomes last and vice versa. For example, if the main hero had 1245 strawberries in the tragedy, he has 5421 of them now. Note that all the leading zeros are omitted. That means if the number ends with a zero, the zero is lost by reversing (e.g. 1200 gives 21). Also note that the reversed number never has any trailing zeros.

ACM needs to calculate with reversed numbers. Your task is to add two reversed numbers and output their reversed sum. Of course, the result is not unique because any particular number is a reversed form of several numbers (e.g. 21 could be 12, 120 or 1200 before reversing). Thus we must assume that no zeros were lost by reversing (e.g. assume that the original number was 12).


The input consists of N cases (equal to about 10000). The first line of the input contains only positive integer N. Then follow the cases. Each case consists of exactly one line with two positive integers separated by space. These are the reversed numbers you are to add.


For each case, print exactly one line containing only one integer - the reversed sum of two reversed numbers. Omit any leading zeros in the output.


Sample input: 
24 1
4358 754
305 794

Sample output:

hide comments
Adithya Abraham Philip: 2015-03-14 06:21:32

To everyone using C struggling with the strrev and atoi functions - they do not belong to the set of standard library functions and as such are not included in gcc.

vaipaliwal: 2015-03-11 19:43:41

1. Does the input have leading zeroes?
eg: 0001

2. If yes, are these to be considered when reversing?
eg 0001 -> 1000 OR 1?

keshavsharma: 2015-03-11 16:53:19

ehh 0.02 time :P

Phạm Bãng Bãng: 2015-01-17 10:44:53

limit ?

sheetal: 2015-01-10 06:52:43

Undefined reference to strrev and itoa. Please help

sheetal: 2015-01-10 06:51:46

I have used string.h and stdlib.h in my code . I am getting this error
prog.c: In function 'main':
prog.c:17:1: warning: implicit declaration of function 'strrev' [-Wimplicit-function-declaration]
prog.c:21:1: warning: implicit declaration of function 'itoa' [-Wimplicit-function-declaration]
/home/X0aXVQ/ccaHMcfd.o: In function `main':
prog.c:(.text.startup+0x79): undefined reference to `strrev'
prog.c:(.text.startup+0x81): undefined reference to `strrev'
prog.c:(.text.startup+0xad): undefined reference to `itoa'
prog.c:(.text.startup+0xb5): undefined reference to `strrev'
collect2: error: ld returned 1 exit status

Ankit: 2014-12-29 17:53:09

stuck with SIGSEGV, I wonder what i am missing... ;-(
[Link Removed]

re(vamsi): please don't post code(link to code in your case) in comments. use forum

Last edit: 2015-02-12 12:39:09
Marcin: 2014-12-13 19:43:49

My code is in java. I reverse first value and second and it is OK. The magic happens when I add those two values (int) and try to reverse them. For first case I have a result for example 133. When I tried to debug it looked like int mod = num%(10^(i+1)) where num is 43 returns 10. Any ideas?

Ashish: 2014-12-01 08:41:18

@Saurabh_P: That was stupid, man. Guys, the answer of 1000 001 should come out to be 101, as you must reverse 001 to read 100 and not just 1.

Ramana Kumar: 2014-11-27 15:22:21

hi all... i am new 2 spoj...
i have tested for the inputs below:
24 1
4358 754
305 794
1000 001


I ran my code on ideone, its successful and the outputs are fine as well..
where could have i gone wrong?
Are there any other validations i m missing? please help.

Added by:adrian
Time limit:5s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Resource:ACM Central European Programming Contest, Prague 1998