// Problem 5.7  recursive pseudocode version
// for y > 0 only
pow (x, y) {
	if ( y == 1 ) return x;
	
	res = pow(x, y >> 1);
	
	if (y & 1)				   // y is odd
		return x * res * res;
		
	else
		return res * res;
	}



// Problem 5.8 solution EPI

string ConvertBase (const string& s, int b1, int b2) {
  bool neg = s.front() == '-';
  
  int x = 0;
  for (size_t i = (neg == true ? 1 : 0); i < s.size(); ++i) {
    x *= b1;
    x += isdigit(s[i]) ? s[i] - '0' : s[i] - 'A' + 10;
    }
    
  string ans;
  while (x) {
    int r = r % b2;
    ans.push_back(r >= 10 ? 'A' + r - 10 : '0' + r);
    x /= b2;
    }
  
  if (ans.empty()) {	// special case: s is 0.
    ans.push_back('0');
    }

  if (neg) {			// s is a negative number
    ans.push_back('-');
    }
  
  reverse(ans.begin(), ans.end() );
  return ans;
  }