結果
| 問題 |
No.3243 Multiplication 8 1
|
| コンテスト | |
| ユーザー |
👑 |
| 提出日時 | 2025-08-22 23:13:20 |
| 言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
AC
|
| 実行時間 | 122 ms / 2,000 ms |
| コード長 | 3,256 bytes |
| コンパイル時間 | 2,122 ms |
| コンパイル使用メモリ | 205,040 KB |
| 実行使用メモリ | 7,716 KB |
| 最終ジャッジ日時 | 2025-08-22 23:14:11 |
| 合計ジャッジ時間 | 2,886 ms |
|
ジャッジサーバーID (参考情報) |
judge1 / judge2 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 1 |
| other | AC * 4 |
ソースコード
/*
[Original Python code — included per AtCoder rule]
MOD = 998244353
def power(A, e):
result = [[1 if i == j else 0 for j in range(len(A))] for i in range(len(A))]
while e:
if e % 2:
result = multiply(result, A)
A = multiply(A, A)
e //= 2
return result
def multiply(A, B):
return [[sum(x * y % MOD for x, y in zip(A_row, B_col)) % MOD for B_col in zip(*B)] for A_row in A]
def multiply_vector(M, v):
return [sum(M[i][j] * v[j] % MOD for j in range(len(v))) % MOD for i in range(len(M))]
X = [
[1, 1, 1, 1, 0, 0, 0], # 1
[1, 1, 1, 1, 0, 0, 0], # -1 ok
[0, 0, 1, 1, 1, 1, 0], # 2
[0, 0, 1, 1, 1, 1, 0], # -2
[1, 0, 0, 0, 1, 1, 1], # 4
[1, 0, 0, 0, 1, 1, 1], # -4
[1, 0, 0, 0, 0, 0, 1], # -1 ng
]
Y = [
[1, 1],
[1, 1],
]
def solve():
N = int(input())
ans = multiply_vector(power(X, N), [1, 0, 0, 0, 0, 0, 0])[0]
ans -= multiply_vector(power(Y, N), [1, 0])[0]
print(ans % MOD)
if __name__ == "__main__":
T = int(input())
for _ in range(T):
solve()
*/
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll MOD = 998244353;
using Mat = vector<vector<ll>>;
static Mat identity(int n) {
Mat I(n, vector<ll>(n, 0));
for (int i = 0; i < n; ++i) I[i][i] = 1;
return I;
}
static Mat multiply(const Mat& A, const Mat& B) {
int n = (int)A.size();
int m = (int)A[0].size(); // also B.size()
int p = (int)B[0].size();
Mat C(n, vector<ll>(p, 0));
for (int i = 0; i < n; ++i) {
for (int k = 0; k < m; ++k) {
ll aik = A[i][k] % MOD;
if (aik == 0) continue;
for (int j = 0; j < p; ++j) {
C[i][j] = (C[i][j] + aik * (B[k][j] % MOD)) % MOD;
}
}
}
return C;
}
static vector<ll> multiply_vector(const Mat& M, const vector<ll>& v) {
int n = (int)M.size();
int m = (int)v.size();
vector<ll> res(n, 0);
for (int i = 0; i < n; ++i) {
ll s = 0;
for (int j = 0; j < m; ++j) {
s = (s + (M[i][j] % MOD) * (v[j] % MOD)) % MOD;
}
res[i] = s;
}
return res;
}
static Mat power(Mat A, long long e) {
int n = (int)A.size();
Mat R = identity(n);
while (e) {
if (e & 1) R = multiply(R, A);
A = multiply(A, A);
e >>= 1;
}
return R;
}
// Constants from the original code
const Mat X = {
{1, 1, 1, 1, 0, 0, 0}, // 1
{1, 1, 1, 1, 0, 0, 0}, // -1 ok
{0, 0, 1, 1, 1, 1, 0}, // 2
{0, 0, 1, 1, 1, 1, 0}, // -2
{1, 0, 0, 0, 1, 1, 1}, // 4
{1, 0, 0, 0, 1, 1, 1}, // -4
{1, 0, 0, 0, 0, 0, 1}, // -1 ng
};
const Mat Y = {
{1, 1},
{1, 1},
};
void solve() {
long long N;
cin >> N;
vector<ll> vX = {1, 0, 0, 0, 0, 0, 0};
vector<ll> vY = {1, 0};
ll a = multiply_vector(power(X, N), vX)[0];
ll b = multiply_vector(power(Y, N), vY)[0];
ll ans = (a - b) % MOD;
if (ans < 0) ans += MOD; // Python's `%` yields non-negative; emulate it
cout << ans << '\n';
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int T;
if (!(cin >> T)) return 0;
while (T--) solve();
return 0;
}