結果
問題 | No.301 サイコロで確率問題 (1) |
ユーザー | heno239 |
提出日時 | 2020-08-23 12:57:42 |
言語 | C++14 (gcc 12.3.0 + boost 1.83.0) |
結果 |
AC
|
実行時間 | 250 ms / 1,000 ms |
コード長 | 3,595 bytes |
コンパイル時間 | 1,147 ms |
コンパイル使用メモリ | 120,280 KB |
実行使用メモリ | 7,680 KB |
最終ジャッジ日時 | 2024-10-15 18:08:40 |
合計ジャッジ時間 | 2,173 ms |
ジャッジサーバーID (参考情報) |
judge4 / judge3 |
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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 | 250 ms
7,680 KB |
testcase_01 | AC | 19 ms
7,680 KB |
ソースコード
#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> #include<array> using namespace std; //#define int long long typedef long long ll; typedef unsigned long long ul; typedef unsigned int ui; constexpr ll mod = 1000000007; const ll INF = mod * mod; 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-12; const ld pi = acosl(-1.0); ll mod_pow(ll x, ll n, ll m) { ll res = 1; while (n) { if (n & 1)res = res * x % m; x = x * x % m; 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, ll 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 = 1 << 18; 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]; } typedef vector<vector<ld>> mat; typedef vector<ld> vec; mat mtmul(mat& A, mat& B) { mat C(A.size(), vec(B[0].size())); rep(i, (int)A.size()) { rep(k, (int)B.size()) { rep(j, (int)B[0].size()) { C[i][j] +=A[i][k] * B[k][j]; } } } return C; } mat mtpow(mat A, ll n) { mat B(A.size(), vec(A.size())); rep(i, (int)A.size()) { B[i][i] = 1; } while (n > 0) { if (n & (ll)1)B = mtmul(B, A); n >>= 1; if (n == 0)break; A = mtmul(A, A); } return B; } void solve() { ll n; cin >> n; /*if (n <= 6) { cout << 6 << "\n"; return; }*/ if (n >= 100000) { ld ans = n + 5.0 / 3.0; cout << ans << "\n"; return; } mat A(7, vec(7,0)); rep1(i, 5) { A[i][i - 1] = 1; } rep(j, 6)A[0][j] = 1 / 6.0; A[0][6] = 1; A[6][6] = 1; A = mtpow(A, n); ld c = 1; rep1(i, 5) { c -= A[0][i]; } ld d = A[0][6]; ld ans = d / c; cout << ans << "\n"; } signed main() { ios::sync_with_stdio(false); cin.tie(0); cout << fixed << setprecision(15); //init_f(); //expr(); int t; cin >> t; rep(i, t) solve(); return 0; }