結果

問題 No.1073 無限すごろく
ユーザー renjyaku_intrenjyaku_int
提出日時 2020-09-10 20:08:47
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 4 ms / 2,000 ms
コード長 2,580 bytes
コンパイル時間 2,861 ms
コンパイル使用メモリ 198,456 KB
最終ジャッジ日時 2025-01-14 09:17:34
ジャッジサーバーID
(参考情報)
judge2 / judge4
このコードへのチャレンジ
(要ログイン)
ファイルパターン 結果
sample AC * 3
other AC * 30
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 <tourist>
#include <bits/stdc++.h>
//#include <atcoder/all>
using namespace std;
//using namespace atcoder;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<ll, ll> p;
const int INF = 1e9;
const ll LINF = ll(1e18);
const int MOD = 1000000007;
const int dx[4] = {0, 1, 0, -1}, dy[4] = {-1, 0, 1, 0};
const int Dx[8] = {0, 1, 1, 1, 0, -1, -1, -1}, Dy[8] = {-1, -1, 0, 1, 1, 1, 0, -1};
#define yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define no cout << "No" << endl
#define NO cout << "NO" << endl
#define rep(i, n) for (int i = 0; i < n; i++)
#define FOR(i, m, n) for (int i = m; i < n; i++)
#define ALL(v) v.begin(), v.end()
#define debug(v) \
cout << #v << ":"; \
for (auto x : v) \
{ \
cout << x << ' '; \
} \
cout << endl;
template <class T>
bool chmax(T &a, const T &b)
{
if (a < b)
{
a = b;
return 1;
}
return 0;
}
template <class T>
bool chmin(T &a, const T &b)
{
if (b < a)
{
a = b;
return 1;
}
return 0;
}
//cout<<fixed<<setprecision(15);15
//-std=c++14
//g++ yarudake.cpp -std=c++17 -I .
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
typedef vector<ll> Array;
typedef vector<Array> Matrix;
ll mod_pow(ll x, ll n, ll mod)
{
ll res = 1LL;
while (n > 0)
{
if (n & 1)
res = res * x % mod;
x = x * x % mod;
n >>= 1;
}
return res;
}
ll mod_inv(ll x, ll mod)
{
return mod_pow(x, mod - 2, mod);
}
Matrix mIdentity(ll n)
{
Matrix A(n, Array(n));
for (int i = 0; i < n; ++i)
A[i][i] = 1;
return A;
}
Matrix mMul(const Matrix &A, const Matrix &B, ll mod)
{
Matrix C(A.size(), Array(B[0].size()));
for (int i = 0; i < C.size(); ++i)
for (int j = 0; j < C[i].size(); ++j)
for (int k = 0; k < A[i].size(); ++k)
(C[i][j] += (A[i][k] % mod) * (B[k][j] % mod)) %= mod;
return C;
}
// O( n^3 log e )
Matrix mPow(const Matrix &A, ll e, ll mod)
{
return e == 0 ? mIdentity(A.size()) : e % 2 == 0 ? mPow(mMul(A, A, mod), e / 2, mod) : mMul(A, mPow(A, e - 1, mod), mod);
}
int main()
{
cin.tie(0);
ios::sync_with_stdio(false);
ll n;
cin >> n;
Matrix x(6, Array(6, 0));
for(int i=0;i<6;i++){
x[0][i]=mod_inv(6,MOD);
}
for(int i=1;i<6;i++){
x[i][i-1]=1;
}
Matrix xn=mPow(x,n,MOD);
cout<<xn[0][0]<<"\n";
}
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