cryptarithm.c~

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/* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
* Description: cross + roads = danger with the extra condition that the digits    *
* assigned to the letters at the beginning of all words (c, rand d) are not equal *
* to 0, and of course such that the resulting algebraic expression is correct.    *
* The output of your program, saved as cryptarithm.c, should be lines of the form *
*                                                                                 *
* cross = ....., roads = ..... and danger = ...... is a solution.                 *
*                                                                                 *
*                                                                                 *
* Written by Suiliang Qiu(student id=z3267009) for COMP9021                       *
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * */ 

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <math.h>

int main(void)
{
    for (int first_text = 12033; first_text <= 98766; ++first_text)
        
        {
            int cross = first_text;
            int roads=1;
            int danger=1;
            int c = cross / 10000;
            int my_bits1 = 1 << c;
            int r = (cross / 1000) % 10;            
            if (r == 0)
                continue;
            if (( my_bits1 >> r) % 2)
                continue;
            my_bits1 +=  1 << r;
            int o= (cross%1000)/100;
            if (( my_bits1 >> o) % 2)
                continue;
            my_bits1 +=  1 << o;
            int s= cross%10;
            if (( my_bits1 >> s) % 2)
                continue;
            my_bits1 +=  1 << s;
            int s2=cross/10%10;
            if(s2!=s)
                continue;
            for (int second_text= 1; second_text <= 98; ++second_text)
                {
                    int my_bits2=my_bits1;
                    int ad = second_text;;
                    int a= ad/10;
                    if (( my_bits2 >> a) % 2)
                        continue;
                    my_bits2 +=  1 << a;
                    int d= ad%10;
                    if(d==0)
                        continue;
                    if (( my_bits2 >> d) % 2)
                        continue;
                    my_bits2 +=  1 << d;
                    roads=s+(ad*10)+o*1000+r*10000;
                    for(int third_text=12;third_text<=987;++third_text)
                        {
                            int my_bits3=my_bits2;
                            int nge=third_text;
                            int n= nge/100;
                            if (( my_bits3 >> n) % 2)
                                continue;
                            my_bits3 +=  1 << n;
                            int g= (nge/10)%10;
                            if (( my_bits3 >> g) % 2)
                                continue;
                            my_bits3 +=  1 << g;
                            int e= nge%10;
                            if (( my_bits3 >> e) % 2)
                                continue;
                            my_bits3 +=  1 << e;
                            danger=r+nge*10+a*10000+d*100000;
                            int result;
                            result =roads+cross;
                            if (danger==result)
                                printf("cross=%d, roads=%d and  danger=%d is a solution\n",cross,roads,danger);
                        }
                    
                }
        }   
    return EXIT_SUCCESS;
}