200+ ACM Problems solution source code

ACMDuring 2003~2007 I was an undergraduate computer science student. In this time for our course material and for programming passion I used to try to solve ACM problems from the well-known University of Valladolid site. I solved arround 203 ACM problems in this site. Today I found in my laptop still the accepted codes are saved. So I think I should share these code. May be these could be useful for someone. I had written 99% of my code using C++ language.

One thing I want to mention these codes are accepted but not the best optimized code. These codes were written by me during my student life so you can’t expect that also.

In my sharing code repository all the file names are the problem number from the University of Valladolid site. In valladolid site author statistics is down, you can’t see anyone’s statistics but fortunately i saved my statistics in 2007 and here it is.

Download Code

About mahmud ahsan

Founder And Lead Programmer at iThinkdiff.net

, , , ,

6 Responses to 200+ ACM Problems solution source code

  1. Joy August 29, 2011 at 9:29 pm #

    Great Share! Cloned in mac, will have a look soon :)

  2. Dictionary of computer science.Engr&technology. December 19, 2011 at 6:04 pm #

    i need a code to open the Dictionary

  3. ACM SLOVE AND SOLUTIONS March 14, 2013 at 4:42 am #

    thankyou brother for great share!

  4. ali November 6, 2013 at 3:13 am #

    Tanx a lot Mahmud
    your codes are really great!!!!!!!!!!
    i promise to use them as well as possible…

  5. walid December 8, 2013 at 3:09 am #

    You are to write a program that evaluates assignment statements. Assume that there
    is one (and only one) statement on each line of the input, and each assignment
    statement is terminated by a semicolon. The following are some of possible
    assignment statements:

    length = 4 +6;
    width=3+12*10;
    area =length *width – b;
    total =100 – y- (width+111*222) /area + x;
    area= area +5 – 3 *4^ ( width – 3*(length+30 ))+100;
    x = area +(total-area/ width ) ;

    Here are some assumptions and rules:

    (1) All variables are 1-10 characters long, and only lowercase letters are
    used.

    (2) All values are integers, and all numbers are in the range of -32768 to
    32767. (Such numbers are stored in 16 bits of memory.)
    (3) / is the integer division operator. So 7/2 = 3 (truncated to integer),
    and 100/7 = 14.
    (4) ^ is the exponential operator. For example, 2^5 = 2*2*2*2*2 = 32.
    (5) An expression may contain both integer literals and variables.

    (6) Any number of spaces may be used to separate different items in a
    statement.
    (7) All statements are 1 – 80 characters long.

    (8) All variables have a default value of 0.

    (9) All statements are valid.
    (10) First line of the input contains a positive integer indicating the
    number of statements to be evaluated.

    Your program should read and evaluate all statements and print a table that shows
    all variables, in alphabetical order and aligned to the left, that appear in the
    input data along with their corresponding final values aligned to right.

    Sample Input

    6
    length = 4 +6;
    width=3+12*10;
    area =length *width – b;
    total =100 – y- (width+111*222) /area + x;
    area= area +5 – 3 *4^ ( width – 3*(length+30 ))+100;
    x = area +(total-area/ width ) ;

    Sample Output

    Variable Value =========================

    area 1143
    b 0
    length 10
    total 80
    width 123
    x 1214
    y 0 please help to solve it

Leave a Reply