#506885 (C++17) No.1102 Remnants

• 問題: No.1102 Remnants
• ユーザー: first_vil
• 提出日時: 2020-07-04 14:58:37
• 言語: C++17 (gcc 12.3.0 + boost 1.83.0)
• 結果: AC
• 実行時間: 54 ms / 2,000 ms
• コード長: 5,078 bytes
• コンパイル時間: 2,444 ms
• コンパイル使用メモリ: 203,548 KB
• 実行使用メモリ: 5,376 KB
• 最終ジャッジ日時: 2024-09-19 02:03:46
• 合計ジャッジ時間: 4,402 ms
• ジャッジサーバーID （参考情報）: judge2 / judge4

入力	結果	実行時間; 実行使用メモリ
testcase_00	AC	2 ms; 5,248 KB
testcase_01	AC	2 ms; 5,376 KB
testcase_02	AC	2 ms; 5,376 KB
testcase_03	AC	2 ms; 5,376 KB
testcase_04	AC	2 ms; 5,376 KB
testcase_05	AC	38 ms; 5,376 KB
testcase_06	AC	9 ms; 5,376 KB
testcase_07	AC	54 ms; 5,376 KB
testcase_08	AC	2 ms; 5,376 KB
testcase_09	AC	4 ms; 5,376 KB
testcase_10	AC	2 ms; 5,376 KB
testcase_11	AC	3 ms; 5,376 KB
testcase_12	AC	4 ms; 5,376 KB
testcase_13	AC	5 ms; 5,376 KB
testcase_14	AC	24 ms; 5,376 KB
testcase_15	AC	19 ms; 5,376 KB
testcase_16	AC	47 ms; 5,376 KB
testcase_17	AC	43 ms; 5,376 KB
testcase_18	AC	47 ms; 5,376 KB
testcase_19	AC	16 ms; 5,376 KB
testcase_20	AC	6 ms; 5,376 KB
testcase_21	AC	28 ms; 5,376 KB
testcase_22	AC	40 ms; 5,376 KB
testcase_23	AC	30 ms; 5,376 KB
testcase_24	AC	22 ms; 5,376 KB
testcase_25	AC	47 ms; 5,376 KB
testcase_26	AC	4 ms; 5,376 KB
testcase_27	AC	13 ms; 5,376 KB

ソースコード

//四則演算 #pragma GCC target("avx")
//並列計算 #pragma GCC optimize("O3")
//条件分岐を減らす #pragma GCC optimize("unroll-loops")
//浮動小数点演算 #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using VI = vector<int>;
using VL = vector<ll>;
using VS = vector<string>;
template<class T> using PQ = priority_queue<T, vector<T>, greater<T>>;
#define FOR(i,a,n) for(int i=(a);i<(n);++i)
#define eFOR(i,a,n) for(int i=(a);i<=(n);++i)
#define rFOR(i,a,n) for(int i=(n)-1;i>=(a);--i)
#define erFOR(i,a,n) for(int i=(n);i>=(a);--i)
#define each(i, a) for(auto &i : a)
#define SORT(a) sort(a.begin(),a.end())
#define rSORT(a) sort(a.rbegin(),a.rend())
#define fSORT(a,f) sort(a.begin(),a.end(),f)
#define all(a) a.begin(),a.end()
#define out(y,x) ((y)<0||h<=(y)||(x)<0||w<=(x))
#define tp(a,i) get<i>(a)
#define line cout << "-----------------------------\n" 
#define ENDL(i,n) ((i) == (n) - 1 ? "\n" : " ")
#define stop system("pause")
constexpr ll INF = 1000000000;
constexpr ll LLINF = 1LL << 60;
constexpr ll mod = 1000000007;
constexpr ll MOD = 998244353;
constexpr ld eps = 1e-10;
constexpr ld pi = 3.1415926535897932;
template<class T>inline bool chmax(T& a, const T& b) { if (a < b) { a = b; return true; }return false; }
template<class T>inline bool chmin(T& a, const T& b) { if (a > b) { a = b; return true; }return false; }
inline void init() { cin.tie(nullptr); cout.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(15); }
template<class T>inline istream& operator>>(istream& is, vector<T>& v) { for (auto& a : v)is >> a; return is; }
template<class T>inline istream& operator>>(istream& is, deque<T>& v) { for (auto& a : v)is >> a; return is; }
template<class T, class U>inline istream& operator>>(istream& is, pair<T, U>& p) { is >> p.first >> p.second; return is; }
template<class T>inline vector<T> vec(size_t a) { return vector<T>(a); }
template<class T>inline vector<T> defvec(T def, size_t a) { return vector<T>(a, def); }
template<class T, class... Ts>inline auto vec(size_t a, Ts... ts) { return vector<decltype(vec<T>(ts...))>(a, vec<T>(ts...)); }
template<class T, class... Ts>inline auto defvec(T def, size_t a, Ts... ts) { return vector<decltype(defvec<T>(def, ts...))>(a, defvec<T>(def, ts...)); }
template<class T>inline void print(const T& a) { cout << a << "\n"; }
template<class T, class... Ts>inline void print(const T& a, const Ts&... ts) { cout << a << " "; print(ts...); }
template<class T>inline void print(const vector<T>& v) { for (int i = 0; i < v.size(); ++i)cout << v[i] << (i == v.size() - 1 ? "\n" : " "); }
template<class T>inline void print(const vector<vector<T>>& v) { for (auto& a : v)print(a); }
inline string reversed(const string& s) { string t = s; reverse(all(t)); return t; }

template<int modulo> struct ModInt {
    int x;

    ModInt() : x(0) {}
    ModInt(long long y) : x(y >= 0 ? y % modulo : (modulo - (-y) % modulo) % modulo) {}

    ModInt& operator+=(const ModInt& p) {
        if ((x += p.x) >= modulo) x -= modulo;
        return *this;
    }
    ModInt& operator-=(const ModInt& p) {
        if ((x += modulo - p.x) >= modulo) x -= modulo;
        return *this;
    }
    ModInt& operator*=(const ModInt& p) {
        x = (int)(1LL * x * p.x % modulo);
        return *this;
    }
    ModInt& operator/=(const ModInt& p) {
        *this *= p.inverse();
        return *this;
    }

    ModInt operator-() const { return ModInt(-x); }
    ModInt operator+(const ModInt& p) const { return ModInt(*this) += p; }
    ModInt operator-(const ModInt& p) const { return ModInt(*this) -= p; }
    ModInt operator*(const ModInt& p) const { return ModInt(*this) *= p; }
    ModInt operator/(const ModInt& p) const { return ModInt(*this) /= p; }

    bool operator==(const ModInt& p) const { return x == p.x; }
    bool operator!=(const ModInt& p) const { return x != p.x; }

    ModInt inverse() const {
        int a = x, b = modulo, u = 1, v = 0, t;
        while (b > 0) {
            t = a / b;
            a -= t * b;
            swap(a, b);
            u -= t * v;
            swap(u, v);
        }
        return ModInt(u);
    }

    ModInt pow(int e) {
        long long a = 1, p = x;
        while (e > 0) {
            if (e % 2 == 0) { p = (p * p) % modulo; e /= 2; }
            else { a = (a * p) % modulo; e--; }
        }
        return ModInt(a);
    }

    friend ostream& operator<<(ostream& os, const ModInt<modulo>& p) {
        return os << p.x;
    }
    friend istream& operator>>(istream& is, ModInt<modulo>& a) {
        long long x;
        is >> x;
        a = ModInt<modulo>(x);
        return (is);
    }
};
using modint = ModInt<mod>;

int main() {
    init();

    int n; modint k; cin >> n >> k;
    vector<modint> a(n); cin >> a;

    vector<modint> ck(n);
    ck[0] = 1;
    FOR(i, 1, n)ck[i] = ck[i - 1] * (k + i) / i;

    modint ans = 0;
    FOR(i, 0, n)ans += a[i] * ck[i] * ck[n - i - 1];
    print(ans);

    return 0;
}