UVA Problem 865 – Substitution Cypher Solution

UVA Problem 865 – Substitution Cypher Solution:

Click here to go to this problem in uva Online Judge.

Solving Technique:

Similar problems to this problem, UVA Problem 10082 – WERTYU Solution and also UVA Problem 11278 – One Handed Typist Solution.

In this problem we are given two strings one of which is plain text and the next one is its substitution. In the plain text string there are characters that are to be mapped to its substitution characters from the substitution string.

Next comes several lines of text in which plain characters that match the required substitution characters should be replaced to its substitution character equivalent. The characters that don’t match the substitution requirement should stay the same.

The hardest part of this problem is newlines. There is a new line after number of cases, so make sure to discard that. Also an empty line marks the end of text inputs for a test case. Last thing is there is a new line between test cases, so for the last output there shouldn’t be an extra new line.


Important:  Be sure to add or print a new line after each output unless otherwise specified. The outputs should match exactly because sometimes even a space character causes the answer to be marked as wrong answer. Please compile with c++ compiler as some of my codes are in c and some in c++.

More Inputs of This Problem on uDebug.



Shar's Birthday:
The birthday is October 6th, but the party will be Saturday,
October 5.  It's my 24th birthday and the first one in some
years for which I've been employed.  Plus, I have new clothes.
So I have cause to celebrate.  More importantly, though,
we've cleaned the house!  The address is 506-D Albert Street.
Extra enticement for CS geeks:  there are several systems in
the house, and the party is conveniently scheduled for 3 hours
after the second CSC programming contest ends (not to mention,
within easy walking distance)!



Sszi'h Brigswzb:
Tsv yrigswzb rh Oxglyvi 6gs, yfg gsv kzigb droo yv Szgfiwzb,
Oxglyvi 5.  Ig'h nb 24gs yrigswzb zmw gsv urihg lmv rm hlnv
bvzih uli dsrxs I'ev yvvm vnkolbvw.  Pofh, I szev mvd xolgsvh.
Sl I szev xzfhv gl xvovyizgv.  Mliv rnkligzmgob, gslfts,
dv'ev xovzmvw gsv slfhv!  Tsv zwwivhh rh 506-D Aoyvig Sgivvg.
Ecgiz vmgrxvnvmg uli CS tvvph:  gsviv ziv hvevizo hbhgvnh rm
gsv slfhv, zmw gsv kzigb rh xlmevmrvmgob hxsvwfovw uli 3 slfih
zugvi gsv hvxlmw CSC kiltiznnrmt xlmgvhg vmwh (mlg gl nvmgrlm,
drgsrm vzhb dzoprmt wrhgzmxv)!


 * Author:    Quickgrid ( Asif Ahmed )
 * Site:      https://quickgrid.wordpress.com
 * Problem:   UVA 865 - Substitution Cypher
 * Technique: Character Mapping, Array Mapping


#define N 256

static char text[N];
static char plaintext[N];
static char substitution[N];

static char newLayout[N];

static char output[N];

int main(){

    int n;

     * Input an integer.
     * discard new line character for enter.
     * discard empty line.
    scanf("%d", &n);

    for(int j = 0; j < n; ++j){

         * Print spaces between test cases.

         * Input plain text and cipher text.

         * Map each character to its ASCII decimal value.
         * In other words map the same character to itself.
         * This is done because not all characters are changed by
         * substitution cipher so rest of the characters map to itself.
        for(int i = 0; i < 256; ++i)
            newLayout[i] = i;

         * Since plain text and cipher text will equal size any of them
         * can be used to check termination condition.
         * Inside the loop map plain text characters to its substitution.
        for(int i = 0; plaintext[i]; ++i)
            newLayout[ plaintext[i] ] = substitution[i];

         * Print substitution string first and plain text as per requirement.

         * Now take input and scramble until a blank line is found.
         * Here i have used string to output each line but character
         * by character output will also work.
        while(gets(text) && strlen(text)){

            memset(output, 0, sizeof output);

            for( int i = 0; i < strlen( text ); ++i )
                output[i] = newLayout[ text[i] ];




    return 0;

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