# include # include # include # include # include # include # include # include # include # include # include # include # include # include # include #include #include #include #include #include using namespace std; typedef long long int ll; const int N = 1000000; const int mod = 1000000007; const int INF = 1 << 30; #define rep(i,n) for(int i=(ll)0;i<(ll)n;++i) #define ALL(x) x.begin(),x.end() #define pp pair #define fi first #define se second ll ppow(ll x, ll n) { ll ans = 1; while (n > 0) { if ((n & 1) == 1)ans = ans + x; x = x * x; n >>= 1; ans %= mod; } return ans; } string YN(bool b) { return(b ? "YES" : "NO"); } string yn(bool b) { return(b ? "Yes" : "No"); } ll n; double v[2000], dp[2000][11]; int main() { cin >> n; cout <<(1<< n) << endl; rep(i, 1 << n)cin >> v[i]; rep(i, 1 << n)dp[i][0] = pow(v[i / 2 * 2 + i % 2],2) / (pow(v[i / 2 * 2],2) + pow(v[i / 2 * 2 + 1],2)); for (int j = 1; j < n; ++j) {//j:回戦 rep(i, 1 << n) {//i:ひと int p; if (i % (1 << j + 1) < (1 << j)) { p = (i / (1 << j) + 1)*(1 << j); for (int k = p; k < p + (1 << j); ++k) { dp[i][j] += (dp[i][j - 1] * dp[k][j - 1] * pow(v[i], 2) / (pow(v[i], 2) + pow(v[k], 2))); } } else { p = p = (i / (1 << j) - 1)*(1 << j); for (int k = p; k < p + (1 << j); ++k) { dp[i][j] += (dp[i][j - 1] * dp[k][j - 1] * pow(v[i], 2) / (pow(v[i], 2) + pow(v[k], 2))); } } } } cout << fixed << setprecision(15) << dp[0][n-1] << endl; return 0; }