結果

問題 No.492 IOI数列
ユーザー 🍮かんプリン
提出日時 2020-06-05 00:25:18
言語 C++11
(gcc 13.3.0)
結果
AC  
実行時間 2 ms / 1,000 ms
コード長 5,330 bytes
コンパイル時間 1,443 ms
コンパイル使用メモリ 166,216 KB
実行使用メモリ 5,248 KB
最終ジャッジ日時 2024-11-30 21:05:38
合計ジャッジ時間 2,470 ms
ジャッジサーバーID
(参考情報)
judge2 / judge1
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ファイルパターン 結果
sample AC * 3
other AC * 19
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

/**
* @FileName a.cpp
* @Author kanpurin
* @Created 2020.06.05 00:25:12
**/
#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
constexpr int MOD = 1e9 + 7;
struct mint {
private:
long long x;
public:
mint(long long x = 0) : x((MOD + x) % MOD) {}
mint(std::string &s) {
long long z = 0;
for (int i = 0; i < s.size(); i++) {
z *= 10;
z += s[i] - '0';
z %= MOD;
}
this->x = z;
}
mint &operator+=(const mint &a) {
if ((x += a.x) >= MOD) x -= MOD;
return *this;
}
mint &operator-=(const mint &a) {
if ((x += MOD - a.x) >= MOD) x -= MOD;
return *this;
}
mint &operator*=(const mint &a) {
(x *= a.x) %= MOD;
return *this;
}
mint &operator/=(const mint &a) {
long long n = MOD - 2;
mint u = 1, b = a;
while (n > 0) {
if (n & 1) {
u *= b;
}
b *= b;
n >>= 1;
}
return *this *= u;
}
mint operator+(const mint &a) const {
mint res(*this);
return res += a;
}
mint operator-(const mint &a) const {
mint res(*this);
return res -= a;
}
mint operator*(const mint &a) const {
mint res(*this);
return res *= a;
}
mint operator/(const mint &a) const {
mint res(*this);
return res /= a;
}
friend std::ostream &operator<<(std::ostream &os, const mint &n) {
return os << n.x;
}
bool operator==(const mint &a) const {
return this->x == a.x;
}
};
template < class T >
struct Matrix {
std::vector< std::vector< T > > A;
Matrix() {}
Matrix(size_t n, size_t m) : A(n, std::vector< T >(m, 0)) {}
Matrix(size_t n) : A(n, std::vector< T >(n, 0)){};
size_t height() const {
return (A.size());
}
size_t width() const {
return (A[0].size());
}
inline const std::vector< T > &operator[](int k) const {
return (A.at(k));
}
inline std::vector< T > &operator[](int k) {
return (A.at(k));
}
static Matrix I(size_t n) {
Matrix mat(n);
for (int i = 0; i < n; i++) mat[i][i] = 1;
return (mat);
}
Matrix &operator+=(const Matrix &B) {
size_t n = height(), m = width();
assert(n == B.height() && m == B.width());
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++) (*this)[i][j] += B[i][j];
return (*this);
}
Matrix &operator-=(const Matrix &B) {
size_t n = height(), m = width();
assert(n == B.height() && m == B.width());
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++) (*this)[i][j] -= B[i][j];
return (*this);
}
Matrix &operator*=(const Matrix &B) {
size_t n = height(), m = B.width(), p = width();
assert(p == B.height());
std::vector< std::vector< T > > C(n, std::vector< T >(m, 0));
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
for (int k = 0; k < p; k++) C[i][j] = (C[i][j] + (*this)[i][k] * B[k][j]);
A.swap(C);
return (*this);
}
Matrix operator+(const Matrix &B) const {
return (Matrix(*this) += B);
}
Matrix operator-(const Matrix &B) const {
return (Matrix(*this) -= B);
}
Matrix operator*(const Matrix &B) const {
return (Matrix(*this) *= B);
}
friend std::ostream &operator<<(std::ostream &os, Matrix &p) {
size_t n = p.height(), m = p.width();
for (int i = 0; i < n; i++) {
os << "[";
for (int j = 0; j < m; j++) {
os << p[i][j] << (j + 1 == m ? "]\n" : ",");
}
}
return (os);
}
T determinant() {
Matrix B(*this);
assert(width() == height());
T ret = 1;
for (int i = 0; i < width(); i++) {
int idx = -1;
for (int j = i; j < width(); j++) {
if (B[j][i] != 0) idx = j;
}
if (idx == -1) return (0);
if (i != idx) {
ret *= -1;
swap(B[i], B[idx]);
}
ret *= B[i][i];
T vv = B[i][i];
for (int j = 0; j < width(); j++) {
B[i][j] /= vv;
}
for (int j = i + 1; j < width(); j++) {
T a = B[j][i];
for (int k = 0; k < width(); k++) {
B[j][k] -= B[i][k] * a;
}
}
}
return (ret);
}
Matrix pow(ll k) const {
auto res = I(A.size());
auto M = *this;
while (k > 0) {
if (k & 1) {
res *= M;
}
M *= M;
k >>= 1;
}
return res;
}
};
int main() {
ll n;
cin >> n;
Matrix< mint > mat(2);
mat.A[0][0] = 100;
mat.A[0][1] = 1;
mat.A[1][0] = 0;
mat.A[1][1] = 1;
auto p = mat.pow(n - 1);
cout << p.A[0][0] + p.A[0][1] << endl;
for (int i = 0; i < n % 11 - 1; i++) {
cout << 10;
}
if (n % 11 == 0) {
cout << 0 << endl;
} else {
cout << 1 << endl;
}
return 0;
}
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