結果
| 問題 | No.344 ある無理数の累乗 |
| ユーザー |
 heno239
|
| 提出日時 | 2020-09-12 11:25:21 |
| 言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
AC
|
| 実行時間 | 5 ms / 2,000 ms |
| コード長 | 3,655 bytes |
| コンパイル時間 | 1,449 ms |
| コンパイル使用メモリ | 120,104 KB |
| 実行使用メモリ | 11,716 KB |
| 最終ジャッジ日時 | 2024-12-31 12:21:40 |
| 合計ジャッジ時間 | 2,615 ms |
|
ジャッジサーバーID （参考情報） |
judge4 / judge1 |
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| ファイルパターン | 結果 |
|---|---|
| other | AC * 30 |
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>//#include<atcoder/dsu>//#include<atcoder/fenwicktree>//#include<atcoder/math>//#include<atcoder/maxflow>using namespace std;//#define int long longtypedef 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-8;const ld pi = acosl(-1.0);ll mod_pow(ll x, ll n, ll m = mod) {if (x >= m)x %= 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 << 19;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];}ll gcd(ll a, ll b) {if (a < b)swap(a, b);while (b) {ll r = a % b; a = b; b = r;}return a;}typedef vector<vector<ll>> mat;typedef vector<ll> 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] = (C[i][j] + (A[i][k] * B[k][j])) % 1000;}}}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);A = mtmul(A, A);n >>= 1;}return B;}void solve(){ld z = 1 + sqrtl(3);int n; cin >> n;mat A = { {1,3},{1,1} };A = mtpow(A, n);ll c = A[0][0]; c *= 2;if (n % 2 == 0)c--;if (c < 0)c += 1000;c %= 1000;cout << c << "\n";}signed main() {ios::sync_with_stdio(false);cin.tie(0);cout << fixed << setprecision(15);//init_f();//init();//expr();//int t; cin >> t; rep(i, t)solve();return 0;}