結果

問題 No.1275 綺麗な式
ユーザー fastmathfastmath
提出日時 2020-10-30 22:55:24
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 3 ms / 2,000 ms
コード長 3,659 bytes
コンパイル時間 1,961 ms
コンパイル使用メモリ 194,852 KB
最終ジャッジ日時 2025-01-15 17:49:55
ジャッジサーバーID
（参考情報） 		judge5 / judge2
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 * 2
other AC * 60
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 # 

#include<bits/stdc++.h>
using namespace std;
#define int long long
#define ii pair <int, int>
#define app push_back
#define all(a) a.begin(), a.end()
#define bp __builtin_popcountll
#define ll long long
#define mp make_pair
#define f first
#define s second
#define Time (double)clock()/CLOCKS_PER_SEC
#define debug(x) std::cout << #x << ": " << x << '\n';


const int MOD = 1e9+7, N = 2;
//need define int long long
int mod(int n) {
    n %= MOD;
    if (n < 0) return n + MOD;
    else return n;
}   
int fp(int a, int p) {
    int ans = 1, c = a;
    for (int i = 0; (1ll << i) <= p; ++i) {
        if ((p >> i) & 1) ans = mod(ans * c);
        c = mod(c * c);
    }   
    return ans;
}   
int dv(int a, int b) { return mod(a * fp(b, MOD - 2)); }

struct M {
ll x;
M (int x_) {
    x = mod(x_);
}   
M () {
    x = 0;
}
M operator + (M y) {
    int ans = x + y.x;
    if (ans >= MOD)
        ans -= MOD;
    return M(ans);
}
M operator - (M y) {
    int ans = x - y.x;
    if (ans < 0)
        ans += MOD;
    return M(ans);            
}   
M operator * (M y) {
    return M(x * y.x % MOD);   
}   
M operator / (M y) {
    return M(x * fp(y.x, MOD - 2) % MOD);
}   
M operator + (int y) {
    return (*this) + M(y);
}
M operator - (int y) {
    return (*this) - M(y);
}   
M operator * (int y) {
    return (*this) * M(y);
}   
M operator / (int y) {
    return (*this) / M(y);
}   
M operator ^ (int p) {
    return M(fp(x, p));
}   
void operator += (M y) {
    *this = *this + y;
}   
void operator -= (M y) {
    *this = *this - y;
}   
void operator *= (M y) {
    *this = *this * y;
}
void operator /= (M y) {
    *this = *this / y;
}   
void operator += (int y) {
    *this = *this + y;
}   
void operator -= (int y) {
    *this = *this - y;
}   
void operator *= (int y) {
    *this = *this * y;
}
void operator /= (int y) {
    *this = *this / y;
}   
void operator ^= (int p) {
    *this = *this ^ p;
}
};  

M f[N], inv[N];
void prec() {
    f[0] = M(1);
    for (int i = 1; i < N; ++i)
        f[i] = f[i - 1] * M(i);
    inv[N - 1] = f[N - 1] ^ (MOD - 2);
    for (int i = N - 2; i >= 0; --i)
        inv[i] = inv[i + 1] * M(i + 1);
}
M C(int n, int k) {
    if (n < k)
        return M(0);
    else
        return f[n] * inv[k] * inv[n - k];
}   

struct Matrix {
M a[N][N];
Matrix () {
    memset(a, 0, sizeof a);
}   
Matrix (bool t) {
    memset(a, 0, sizeof a);
    for (int i = 0; i < N; ++i)
        a[i][i] = t;
}   
M * operator [] (int i) {
    return a[i];
}   
Matrix operator * (Matrix b) {
    Matrix ans;
    for (int i = 0; i < N; ++i)
        for (int j = 0; j < N; ++j)
            for (int k = 0; k < N; ++k)
                ans[i][j] += a[i][k] * b[k][j];
    return ans;
}   
void operator *= (Matrix b) {
    *this = *this * b;
}   
void print() {
    for (int i = 0; i < N; ++i) {
        for (int j = 0; j < N; ++j) 
            cout << a[i][j].x << ' ';
        cout << '\n';
    }   
    cout << '\n';
}
Matrix operator ^ (int p) { 
    Matrix cur = *this;
    Matrix ans(1);
    for (int i = 0; (1ll << i) <= p; ++i) {
        if ((p >> i) & 1) 
            ans *= cur;
        cur *= cur;
    }       
    return ans;
}   
void operator ^= (int p) {
    *this = *this ^ p;
}   
};

signed main() {
    #ifdef HOME
    freopen("input.txt", "r", stdin);
    #else
    #define endl '\n'
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    cout.setf(ios::fixed); cout.precision(20); 
    #endif

    int A, B, n;
    cin >> A >> B >> n;

    Matrix a;
    a[0][0] = A;
    a[1][1] = A;
    a[0][1] = B;
    a[1][0] = 1;

    a ^= n;

    M ans = a[0][0] + a[1][1];
    cout << ans.x << endl;            
}
0