結果

問題 No.535 自然数の収納方法
ユーザー はまやんはまやんはまやんはまやん
提出日時 2017-06-24 01:15:00
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 48 ms / 2,000 ms
コード長 3,278 bytes
コンパイル時間 1,535 ms
コンパイル使用メモリ 169,044 KB
実行使用メモリ 35,328 KB
最終ジャッジ日時 2024-10-03 04:09:13
合計ジャッジ時間 3,170 ms
ジャッジサーバーID
(参考情報)
judge5 / judge1
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 24 ms
35,144 KB
testcase_01 AC 24 ms
35,272 KB
testcase_02 AC 24 ms
35,272 KB
testcase_03 AC 24 ms
35,272 KB
testcase_04 AC 24 ms
35,144 KB
testcase_05 AC 26 ms
35,140 KB
testcase_06 AC 26 ms
35,268 KB
testcase_07 AC 31 ms
35,144 KB
testcase_08 AC 39 ms
35,200 KB
testcase_09 AC 41 ms
35,328 KB
testcase_10 AC 44 ms
35,200 KB
testcase_11 AC 29 ms
35,328 KB
testcase_12 AC 34 ms
35,200 KB
testcase_13 AC 26 ms
35,200 KB
testcase_14 AC 32 ms
35,276 KB
testcase_15 AC 45 ms
35,200 KB
testcase_16 AC 39 ms
35,144 KB
testcase_17 AC 47 ms
35,144 KB
testcase_18 AC 26 ms
35,200 KB
testcase_19 AC 48 ms
35,200 KB
testcase_20 AC 47 ms
35,144 KB
testcase_21 AC 47 ms
35,144 KB
testcase_22 AC 48 ms
35,276 KB
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>
#define rep(i,a,b) for(int i=a;i<b;i++)
#pragma GCC optimize ("-O3")
using namespace std; void _main(); int main() { cin.tie(0); ios::sync_with_stdio(false); _main(); }
//---------------------------------------------------------------------------------------------------
template<int MOD> struct ModInt {
    static const int Mod = MOD; unsigned x; ModInt() : x(0) { }
    ModInt(signed sig) { x = sig < 0 ? sig % MOD + MOD : sig % MOD; }
    ModInt(signed long long sig) { x = sig < 0 ? sig % MOD + MOD : sig % MOD; }
    int get() const { return (int)x; }
    ModInt &operator+=(ModInt that) { if ((x += that.x) >= MOD) x -= MOD; return *this; }
    ModInt &operator-=(ModInt that) { if ((x += MOD - that.x) >= MOD) x -= MOD; return *this; }
    ModInt &operator*=(ModInt that) { x = (unsigned long long)x * that.x % MOD; return *this; }
    ModInt &operator/=(ModInt that) { return *this *= that.inverse(); }
    ModInt operator+(ModInt that) const { return ModInt(*this) += that; }
    ModInt operator-(ModInt that) const { return ModInt(*this) -= that; }
    ModInt operator*(ModInt that) const { return ModInt(*this) *= that; }
    ModInt operator/(ModInt that) const { return ModInt(*this) /= that; }
    ModInt inverse() const { long long a = x, b = MOD, u = 1, v = 0;
        while (b) { long long t = a / b; a -= t * b; std::swap(a, b); u -= t * v; std::swap(u, v); }
        return ModInt(u); }
    bool operator==(ModInt that) const { return x == that.x; }
    bool operator!=(ModInt that) const { return x != that.x; }
    ModInt operator-() const { ModInt t; t.x = x == 0 ? 0 : Mod - x; return t; }
};
template<int MOD> ModInt<MOD> operator^(ModInt<MOD> a, unsigned long long k) {
    ModInt<MOD> r = 1; while (k) { if (k & 1) r *= a; a *= a; k >>= 1; } return r; }
typedef ModInt<1000000007> mint;
/*---------------------------------------------------------------------------------------------------
            ∧_∧  
      ∧_∧  (´<_` )  Welcome to My Coding Space!
     ( ´_ゝ`) /  ⌒i     
    /   \     | |     
    /   / ̄ ̄ ̄ ̄/  |  
  __(__ニつ/     _/ .| .|____  
     \/____/ (u ⊃  
---------------------------------------------------------------------------------------------------*/



int N;
//---------------------------------------------------------------------------------------------------
mint dp[2020][2020];
mint sm[2020][2020];
void dodp() {
    rep(i, 1, N) {
        sm[i - 1][0] = 0;
        rep(j, 1, N + 1) sm[i - 1][j] = sm[i - 1][j - 1] + dp[i - 1][j];
        rep(j, 1, N + 1) dp[i][j] = sm[i - 1][min(N, j + max(1, i - 1) - 1)];
    }
}
//---------------------------------------------------------------------------------------------------
void _main() {
    cin >> N;
    
    // dp1: An = any -> ... -> An-1 = any
    rep(i, 0, 2020) rep(j, 0, 2020) dp[i][j] = 0;
    rep(j, 1, N + 1) dp[0][j] = 1;
    dodp();

    mint ans = 0;
    rep(j, 1, N + 1) ans += dp[N - 1][j];

    // dp2: An = 1 -> ... -> An-1 = N
    rep(i, 0, 2020) rep(j, 0, 2020) dp[i][j] = 0;
    dp[0][1] = 1;
    dodp();

    ans -= dp[N - 1][N];

    cout << ans.get() << endl;
}
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