結果

問題 No.194 フィボナッチ数列の理解(1)
ユーザー HachimoriHachimori
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
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 #
プレゼンテーションモードにする

#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));
else
return 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;
}
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