#include using namespace std; /* #challenge The digits sum of a multiple of 3, gives another multiple of 3. It's a known fact. Do a program that sums a number's digits, sums the result's digits again and again until the result is only one digit long. If that resulting digit is 3,6 or 9, your program outputs "multiple of 3". example: n=1239 1+2+3+9=15 1+5=6 output "multiple of 3" Your program should be able to take big numbers like 12345678901234567890123456789012345678901234567890111222333837366363663636626626516352562718 and tell if is a multiple of 3. */ int main() { /* The code checks input as well, eliminating those if (c<'0') ... state == 3 indicates wrong character in the input flow */ short** stateTable = new short*[4]; for(auto i = 0; i < 4; i++) { stateTable[i] = new short[256]; for (auto j= 0; j<256; j++) { stateTable[i][j] = 3; } } short t = 0; for(auto i = 0; i < 3; ++i) { for (auto j= '0'; j<='9'; j++) { stateTable[i][j] = t++; t %= 3; } //allow white spaces stateTable[i][' '] = i; stateTable[i]['\r'] = i; stateTable[i]['\n'] = i; stateTable[i]['\t'] = i; } // Note the first letter l (small L) instead of digit 1 char num[] = "l234567890123456789012 345678901234567890123456789011122 2333837366363663636626626516352562718"; short state = 0; char *p = num; while (*p) state = stateTable[state][(unsigned char)(*p++)]; cout << "state=" << state << endl; if (state == 3) cout << "wrong input" << endl; else if (state) cout << "given number is NOT a multiple of 3" << endl; else cout << "given number IS a multiple of 3" << endl; return state; // ;) }

is mul of 3 v3