#include #include #include #include #include #include #include #include #include #include #include #include #define rep(i,n) for(int i=0;i<(int)(n);++i) #define F first #define S second #define SZ(a) (int)((a).size()) #define sz(a) SZ(a) #define pb(a) push_back(a) #define mp(a,b) make_pair(a,b) #define ALL(a) (a).begin(),(a).end() using namespace std; typedef long long ll; typedef pair PI; typedef unsigned long long ull; #define PR(...) do{cerr << "line : " << __LINE__ << endl; pr(#__VA_ARGS__, __VA_ARGS__);}while(0); template void pr(const string& name, T t){ cerr << name << ": " << t << endl; } template void pr(const string& names, T t, Types ... rest) { auto comma_pos = names.find(','); cerr << names.substr(0, comma_pos) << ": " << t << ", "; auto next_name_pos = names.find_first_not_of(" \t\n", comma_pos + 1); pr(string(names, next_name_pos), rest ...); } template ostream& operator<< (ostream& o, const pair& v){return o << "(" << v.F << ", " << v.S << ")";} template ostream& operator<< (ostream& o, const vector& v){o << "{";rep(i,SZ(v)) o << (i?", ":"") << v[i];return o << "}";} #define endl '\n' int k; double memo[1100000]; bool vis[1100000]; double pro[7]; double rec(int cur){ if(cur >= k) return 0; if(vis[cur]) return memo[cur]; vis[cur] = 1; double& ret = memo[cur] = 1; for(int i = 1; i <= 6; ++i) ret += rec(cur+i) * pro[i]; return ret; } /* 1*(1-x) + x*2 = 1.0833333333333333 1*(1-x1-x2) + x1*x1*3 + x1 * (1-x1) * 2 + x2 * (1-x1) * 2 = 1.2569444444444444 */ int main(int argc, char *argv[]) { //rep(i,6) pro[i+1] = 1./6; double prd[] = {1.0000000000000000,1.0833333333333333,1.2569444444444444,1.5353009259259260,1.6915991512345676,2.0513639724794235}; rep(i,5){ double sum = 0; rep(j,i) sum += pro[j+1]; double low = 0, up = 1-sum; //cout << sum << endl; k=i+2; rep(k,100){ double mid = (low+up)/2; pro[i+1] = mid; pro[i+2] = 1-sum-mid; memset(vis,0,sizeof(vis)); //cout << rec(0) << endl; if(rec(0) < prd[i+1]) low = mid; else up = mid; } //cout << low << endl; pro[i+1] = low; //cout << rec(0) << endl; } //return 0; int t; cin >> t; rep(i,t){ //int k; cin >> k; while(k<=1e6); memset(vis,0,sizeof(vis)); printf("%.18f\n",rec(0)); } return 0; }