結果

問題 No.76 回数の期待値で練習
ユーザー heno239heno239
提出日時 2020-02-11 12:55:14
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 1,064 ms / 5,000 ms
コード長 4,158 bytes
コンパイル時間 1,023 ms
コンパイル使用メモリ 115,780 KB
実行使用メモリ 33,180 KB
最終ジャッジ日時 2024-10-01 07:16:47
合計ジャッジ時間 2,374 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge4
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 2 ms
6,816 KB
testcase_01 AC 1,064 ms
33,180 KB
権限があれば一括ダウンロードができます

ソースコード

diff # 

#include<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
#include<cassert>
#include<complex>
#include<numeric>
using namespace std;

//#define int long long
typedef long long ll;

typedef unsigned long long ul;
typedef unsigned int ui;
const ll mod = 1000000007;
const ll INF = (1e+18) + 7;
typedef pair<int, int>P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
#define all(v) (v).begin(),(v).end()
typedef pair<ll, ll> LP;
typedef long double ld;
typedef pair<ld, ld> LDP;
const ld eps = 1e-6;
const ld pi = acos(-1.0);
//typedef vector<vector<ll>> mat;
typedef vector<int> vec;

void debug(vector<int> &v) {
	rep(i, v.size()) {
		if (i > 0)cout << " ";
		cout << v[i];
	}
	cout << endl;
}

ll mod_pow(ll a, ll n) {
	ll res = 1;
	while (n) {
		if (n & 1)res = res * a%mod;
		a = a * a%mod; n >>= 1;
	}
	return res;
}

struct modint {
	ll n;
	modint() :n(0) { ; }
	modint(ll m) :n(m) {
		if (n >= mod)n %= mod;
		else if (n < 0)n = (n%mod + mod) % mod;
	}
	operator int() { return n; }
};
bool operator==(modint a, modint b) { return a.n == b.n; }
modint operator+=(modint &a, modint b) { a.n += b.n; if (a.n >= mod)a.n -= mod; return a; }
modint operator-=(modint &a, modint b) { a.n -= b.n; if (a.n < 0)a.n += mod; return a; }
modint operator*=(modint &a, modint b) { a.n = ((ll)a.n*b.n) % mod; return a; }
modint operator+(modint a, modint b) { return a += b; }
modint operator-(modint a, modint b) { return a -= b; }
modint operator*(modint a, modint b) { return a *= b; }
modint operator^(modint a, int n) {
	if (n == 0)return modint(1);
	modint res = (a*a) ^ (n / 2);
	if (n % 2)res = res * a;
	return res;
}

ll inv(ll a, ll p) {
	return (a == 1 ? 1 : (1 - p * inv(p%a, a)) / a + p);
}
modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }

const int max_n = 2050;
modint fact[max_n], factinv[max_n];
void init_f() {
	fact[0] = modint(1);
	for (int i = 0; i < max_n - 1; i++) {
		fact[i + 1] = fact[i] * modint(i + 1);
	}
	factinv[max_n - 1] = modint(1) / fact[max_n - 1];
	for (int i = max_n - 2; i >= 0; i--) {
		factinv[i] = factinv[i + 1] * modint(i + 1);
	}
}
modint comb(int a, int b) {
	if (a < 0 || b < 0 || a < b)return 0;
	return fact[a] * factinv[b] * factinv[a - b];
}

struct uf {
private:
	vector<int> par, ran;
public:
	uf(int n) {
		par.resize(n, 0);
		ran.resize(n, 0);
		rep(i, n) {
			par[i] = i;
		}
	}
	int find(int x) {
		if (par[x] == x)return x;
		else return par[x] = find(par[x]);
	}
	void unite(int x, int y) {
		x = find(x), y = find(y);
		if (x == y)return;
		if (ran[x] < ran[y]) {
			par[x] = y;
		}
		else {
			par[y] = x;
			if (ran[x] == ran[y])ran[x]++;
		}
	}
	bool same(int x, int y) {
		return find(x) == find(y);
	}
};
int dx[4] = { 0,1,0,-1 };
int dy[4] = { 1,0,-1,0 };


ld p[7];
void solve() {
	/*vector<ld> a(7);
	rep1(i, 6)cin >> a[i];
	rep1(i, 6) {
		vector<ld> dp(7);
		for (int j = i - 1; j > 0; j--) {
			dp[j] = 1;
			rep1(ad, 6) {
				if (j + ad >= i)continue;
				dp[j] += dp[j + ad]*p[ad];
			}
		}
		ld sum = 1;
		for (int ad = 1; ad < i - 1; ad++) {
			sum += dp[ad]*p[ad];
		}
		p[i - 1] = a[i] - sum;
	}
	p[6] = 1;
	rep1(i, 5)p[6] -= p[i];
	rep1(i, 6)cout << p[i] << endl;*/

	p[1] = p[4] = 1 / 12.0;
	p[2] = p[6] = 1 / 6.0;
	p[3] = p[5] = 1 / 4.0;


	int t; cin >> t;
	rep(aa, t) {
		int n; cin >> n;
		vector<ld> dp(n);
		per(i, n) {
			dp[i] = 1;
			rep1(ad, 6) {
				if (i + ad >= n)continue;
				dp[i] += dp[i + ad] * p[ad];
			}
		}
		cout << dp[0] << endl;
	}
}


signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout << fixed << setprecision(10);
	//init_f(); //init();
	//init();
	//int t; cin >> t; rep(i, t)solve();
	//while(cin>>w>>h,w)solve();
	solve();
	stop
		return 0;
}
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