結果
問題 | No.194 フィボナッチ数列の理解(1) |
ユーザー | Hachimori |
提出日時 | 2015-04-26 23:37:27 |
言語 | C++11 (gcc 13.3.0) |
結果 |
AC
|
実行時間 | 18 ms / 5,000 ms |
コード長 | 2,572 bytes |
コンパイル時間 | 822 ms |
コンパイル使用メモリ | 65,676 KB |
実行使用メモリ | 7,252 KB |
最終ジャッジ日時 | 2024-07-05 02:44:26 |
合計ジャッジ時間 | 2,235 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge1 |
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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<vector>using namespace std;typedef vector<int> VI;typedef vector<VI> VII;const int BUF_LARGE = 10005;const int BUF_SMALL = 35;const int RESULT = 1000005;const int MOD = 1000000007;int N;long long K;int A[BUF_LARGE];void read() {cin >> N >> K;for (int i = 0; i < N; ++i)cin >> A[i];}VII matmul(const VII &a, const VII &b) {VII ret = VII(a.size(), VI(b[0].size()));for (int i = 0; i < a.size(); ++i)for (int j = 0; j < b[0].size(); ++j)for (int k = 0; k < a[0].size(); ++k)ret[i][j] = (ret[i][j] + 1LL * a[i][k] * b[k][j]) % MOD;return ret;}VII matpow(VII &p, long long e) {if (e == 0) {VII vii = VII(p.size(), VI(p.size()));for (int i = 0; i < p.size(); ++i) vii[i][i] = 1;return vii;}VII t = matpow(p, e / 2);if (e & 1)return matmul(p, matmul(t, t));elsereturn matmul(t, t);}void work() {if (N <= 30) {// calc F(K)VII p = VII(N, VI(N));for (int i = 0; i < N; ++i)p[0][i] = 1;for (int r = 1, c = 0; r < N; ++r, ++c)p[r][c] = 1;VII ret = matpow(p, K - N);int ans = 0;for (int i = 0; i < N; ++i)ans = (ans + 1LL * ret[0][i] * A[N - i - 1]) % MOD;cout << ans << ' ';// calc S(K)p = VII(N + 1, VI(N + 1));p[0][0] = 2;p[0][N] = MOD - 1;for (int r = 1, c = 0; r < N + 1; ++r, ++c)p[r][c] = 1;ret = matpow(p, K - N);int sum = 0;for (int i = 0; i < N; ++i)sum = (sum + A[i]) % MOD;ans = 0;for (int i = 0; i < N; ++i) {ans = (ans + 1LL * ret[0][i] * sum) % MOD;sum = (sum + MOD - A[N - 1 - i]) % MOD;}cout << ans << endl;}else {static int result[RESULT];for (int i = 0; i < N; ++i)result[i] = A[i];int sum = 0;int sk = 0;for (int i = 0; i < N; ++i) {sum = (sum + A[i]) % MOD;sk = (sk + A[i]) % MOD;}for (int i = N; i < K; ++i) {result[i] = sum;sk = (sk + sum) % MOD;sum = (1LL * sum + sum + MOD - result[i - N]) % MOD;}cout << result[K - 1] << ' ' << sk << endl;}}int main() {read();work();return 0;}