3 const char *ls = "1111100111111001110010101101101011111000100011111110100110011110" // A-D
4 "1111111010001111111110001110100011111000100111111001100111111001" // E-H
5 "1111010001001111111100011001111110111100101010011000100010001111" // I-L
6 "1111101110011001110110111001100111111001100111111111100111111000" // M-P
7 "1111100110111111111110011111101011111000111101111111010001000100" // Q-T
8 "1001100110011111100110011001011010011001101111111001011001101001" // U-X
9 "10011001111101001111001001001111"; // Y-Z
12 "f9f935b5f11f7997f71ff171f19f99f9f22ff89fd359111ffd99bd99f99ff9f1f9dff9f5f1fef222999f999699df966999f2f42f";
14 const wchar_t *uc = L"ùù5µñ
\1fy
\97÷
\1fñqñ
\9f\99ùò/ø
\9fÓY
\11\1fý
\99½
\99ù
\9fùñùßùõñþò\"
\99\9f\99\96\99ß
\96i
\99òô/";
18 for (int o = 0; o < 4; o++)
20 int x = hs[(c - 'a') * 4 + o];
21 x -= x >= 'a' ? ('a' - 10) : '0';
22 for (int i = 0; i < 4; i++)
23 printf("%c%s", (x & (1 << i)) > 0 ? 'X' : ' ', (i + 1) % 4 == 0 ? "\n" : "");
28 for (int o = 0; o < 2; o++)
30 int x = uc[(c - 'a') * 2 + o];
31 for (int i = 0; i < 8; i++)
32 printf("%c%s", (x & (1 << i)) > 0 ? 'X' : ' ', (i + 1) % 4 == 0 ? "\n" : "");
38 for (int i = 0; i < strlen(ls) / 4; i++)
41 for (int j = 0; j < 4; j++)
43 char c = ls[i * 4 + (3 - j)];
45 num += (1 << (3 - j));
53 for (int i = 0; i < strlen(ls) / 8; i++)
56 for (int j = 0; j < 8; j++)
58 char c = ls[i * 8 + (7 - j)];
60 num += (1 << (7 - j));