r[]={6,16644,488807,16391,1010519,772}; p(n){n>9&&p(n/10);printf("### \0 # \0# \0# # "+5*(r[*r]>>n%10*2&3));} main(n){for(scanf("%d",&n);--*r;)p(n),puts("");} /* * +=========-- PRETTY PRINTER --=========+ * | Program takes long number from 'stdin' | * | and writes it's prettified version | * | to the stdout | * | | * | This is a code golf program, | * | so it contains least amount | * | of symbols possible | * | | * | Made for 'One Lone Coder' | * | discord: WhwHUMV | * | | * | Copyright (c) Ermishin Michael 2017 | * +========================================+ * * * Output digits format: * * 0 1 2 3 4 5 6 7 8 9 * ### # ### ### # # ### ### ### ### ### * # # # # # # # # # # # # # # * # # # ### ### ### ### ### # ### ### * # # # # # # # # # # # # # * ### # ### ### # ### ### # ### ### */ /* * ===--- HOW DOES IT WORKS? ---=== * * * ==-- ARRAY 'r' AND NUMBER REPRESENTATION --== * * Array 'r' stores 6 values: * 0: current line * 1: line 1 digits representation * 2: line 2 digits representation * 3: line 3 digits representation * 4: line 4 digits representation * 5: line 5 digits representation * * Digits representation consists * of 20 bits (10 bit pairs) * 1 pair for each digit * * Pair indicates the highlight state * for given digit * * Name Pair Look Type * F 00 ### full * L 01 #.. left * R 10 ..# right * M 11 #.# 2-sided * * Representation data for each line * 0 1 2 3 4 5 6 7 8 9 * 1: F R F F M F F F F F * 2: M R R R M L L R M M * 3: M R F F F F F R F F * 4: M R L R R R M R M R * 5: F R F F R F F R F R * * Representation in quaternary number system * 0123456789 * 1: 0200300000 * 2: 3222311233 * 3: 3200000200 * 4: 3212223232 * 5: 0200200202 * * Representation in binary number system * 1: 00100000110000000000 * 2: 11101010110101101111 * 3: 11100000000000100000 * 4: 11100110101011101110 * 5: 00100000100000100000 * * For faster access using bit-shifting * reverse all these numbers * 1: 00000000001100000100 * 2: 11110110101101010111 * 3: 00000100000000000111 * 4: 01110111010101100111 * 5: 00000100000100000100 * * Then use their decimal representation * to reduce program size * 1: 772 * 2: 1010519 * 3: 16391 * 4: 488807 * 5: 16644 * * * ==-- MAIN FUNCTION --== * * Reading the number from stdin * is placed into loop's init place * in order to remove 1 unneeded semicolon * * --*r decrements current line by 1 * and assumes that it isn't zero(0) * * p(n) prints one line of n's representation * * , operator to evaluate multiple statements * used to get rid of braces '{}' * and semicolons ';' * * After that move to the next line * * * ==-- REPRESENTATION LINE PRINTER --== * * method 'p' takes 1 parameter - number to print * it calls itself recursively * to print each digit in the right order * * n > 9 && p(n / 10) * calls function to print the next digit * only if it is possible * to prevent infinite recursion * * * "### \0 # \0# \0# # " * substring mapping * might be represented as list * 0: ###_ full * 5: __#_ right * 10: #___ left * 15: #_#_ 2-sided * with offsets that are multiple of 5 * left & right values are swapped * because binary representation is reversed * * *r - gets current line * r[*r] - gets digits representation for current line * * n % 10 - last digit of a number * * r[*r] >> n % 10 * 2 * shifts binary representation * to match current digit * * ... & 3 * & operation has least priority in the group * it is used to get state pair (2 bits) * * 5 * ... * convert state index to the right offset * * ... + ... * get n-th state representation * doesn't output whole string * because of \0 delimiters */ /* ===--- VERSION INFO ---=== * * * =- Version 1 [244 bytes] -= * Initial version * * #include * using std::cout; * void p(int n,int l){if(n/10)p(n/10,l);cout<<"### \0 # \0# \0# # "+5*(((new int[5]{0x4104,0x77567,0x04007,0xF6B57,0x304})[l]>>n%10*2)&3);} * int main(){int n;std::cin>>n;for(int i=5;i;) p(n,--i),cout<<'\n';} * * * =- Version 2 [209 bytes] -= * Moved array declaration * Removed 'line' parameter * Changed hex numbers to decimal * Moved reading into for loop init * * #include * long r[]{16644,488807,16391,1010519,772},i=5,n; * int p(long n){if(n/10)p(n/10);std::cout<<"### \0 # \0# \0# # "+5*(r[i]>>n%10*2&3);} * int main(){for(std::cin>>n;i--;)p(n),std::cout<<'\n';} * * * =- Version 3 [204 bytes] -= * Replaced if statement by and(&&) operator * Moved current line into data holder * * #include * long r[]{6,16644,488807,16391,1010519,772},n; * int p(long n){n>9&&p(n/10);std::cout<<"### \0 # \0# \0# # "+5*(r[*r]>>n%10*2&3);} * int main(){for(std::cin>>n;--*r;)p(n),std::cout<<'\n';} * * * =- Version 4 [202 bytes] -= * Changed #include to #import * Removed space after #import * * * =- Version 5 [187 bytes] -= * imported cstdio instead of iostream * used scanf & printf instead of cin & cout * used puts("") instead of printf("\n") * changed long to int * * * =- Version 6 [184 bytes] -= * imported ios instead of cstdio * * * =- Version 7 [159 bytes] -= * removed import * used implicit declarations */

Number pretty printer [GOLF]