結果
問題 | No.194 フィボナッチ数列の理解(1) |
ユーザー | heno239 |
提出日時 | 2020-08-09 17:00:48 |
言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 21 ms / 5,000 ms |
コード長 | 3,967 bytes |
コンパイル時間 | 1,462 ms |
コンパイル使用メモリ | 123,644 KB |
実行使用メモリ | 26,916 KB |
最終ジャッジ日時 | 2024-10-05 00:49:37 |
合計ジャッジ時間 | 2,887 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge3 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 37 |
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>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 double ld;typedef pair<ld, ld> LDP;const ld eps = 1e-12;const ld pi = acos(-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 << 17;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<modint>> mat;typedef vector<modint> 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);A = mtmul(A, A);n >>= 1;}return B;}void solve() {int n;ll k; cin >> n >> k;vector<modint> a(n);rep(i, n) {ll in; cin >> in; a[i] = in;}if (n <= 30) {mat A(n+1, vector<modint>(n + 1, 0));rep(j, n) {A[0][j] = 1;}rep(j, n + 1)A[n][j] = 1;rep1(j, n - 1) {A[j][j - 1] = 1;}mat B = mtpow(A, k - n);reverse(all(a));modint ans1 = 0, ans2 = 0;modint sum = 0; rep(i, n)sum += a[i];a.push_back(sum);rep(i, n+1) {ans1 += B[0][i] * a[i];ans2 += B[n][i] * a[i];}cout << ans1 << " " << ans2 << "\n";}else {vector<modint> ra(n + 1);rep(i, n) {ra[i + 1] = ra[i] + a[i];}while (a.size() < k) {modint sum = ra[a.size()] - ra[a.size() - n];ra.push_back(sum + ra.back());a.push_back(sum);}cout << a.back() << " " << ra.back() << "\n";}}signed main() {ios::sync_with_stdio(false);cin.tie(0);//cout << fixed << setprecision(10);//init_f();//expr();//int t; cin >> t; rep(i, t)solve();return 0;}