結果

問題 No.685 Logical Operations
ユーザー 👑 はまやんはまやんはまやんはまやん
提出日時 2018-05-12 00:05:16
言語 C++14
(gcc 13.2.0 + boost 1.83.0)
結果
AC  
実行時間 2 ms / 2,000 ms
コード長 4,884 bytes
コンパイル時間 1,474 ms
コンパイル使用メモリ 168,496 KB
実行使用メモリ 4,384 KB
最終ジャッジ日時 2023-09-10 17:57:38
合計ジャッジ時間 2,513 ms
ジャッジサーバーID
(参考情報)
judge12 / judge13
このコードへのチャレンジ
(要ログイン)

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 1 ms
4,376 KB
testcase_01 AC 1 ms
4,376 KB
testcase_02 AC 2 ms
4,380 KB
testcase_03 AC 1 ms
4,376 KB
testcase_04 AC 1 ms
4,380 KB
testcase_05 AC 2 ms
4,376 KB
testcase_06 AC 2 ms
4,376 KB
testcase_07 AC 2 ms
4,384 KB
testcase_08 AC 1 ms
4,380 KB
testcase_09 AC 2 ms
4,380 KB
testcase_10 AC 1 ms
4,380 KB
testcase_11 AC 2 ms
4,376 KB
testcase_12 AC 2 ms
4,380 KB
testcase_13 AC 1 ms
4,380 KB
testcase_14 AC 2 ms
4,380 KB
testcase_15 AC 1 ms
4,376 KB
testcase_16 AC 1 ms
4,376 KB
testcase_17 AC 2 ms
4,380 KB
testcase_18 AC 1 ms
4,376 KB
testcase_19 AC 1 ms
4,376 KB
testcase_20 AC 2 ms
4,376 KB
testcase_21 AC 1 ms
4,384 KB
testcase_22 AC 2 ms
4,376 KB
testcase_23 AC 1 ms
4,376 KB
testcase_24 AC 1 ms
4,376 KB
testcase_25 AC 2 ms
4,376 KB
testcase_26 AC 1 ms
4,380 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++)
#define rrep(i,a,b) for(int i=a;i>=b;i--)
#define fore(i,a) for(auto &i:a)
#define all(x) (x).begin(),(x).end()
#pragma GCC optimize ("-O3")
using namespace std; void _main(); int main() { cin.tie(0); ios::sync_with_stdio(false); _main(); }
typedef long long ll; const int inf = INT_MAX / 2; const ll infl = 1LL << 60;
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; }
//---------------------------------------------------------------------------------------------------
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> ostream& operator<<(ostream& st, const ModInt<MOD> a) { st << a.get(); return st; };
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 ⊃  
---------------------------------------------------------------------------------------------------*/






ll N;
//---------------------------------------------------------------------------------------------------
// state1
// 0:x=y=N
// 1:x<y=N
// 2:x=y<N
// 3:x<y<N

// state2
// 0:and=xor
// 1:and<xor

// state3
// 0:xor=or
// 1:xor<or
mint memo[70][4][2][2];
int vis[70][4][2][2];
mint f(int dig, int state1, int state2, int state3) {
    if (dig < 0) {
        if (state2 == 1 and state3 == 1) return 1;
        else return 0;
    }

    if (vis[dig][state1][state2][state3]) return memo[dig][state1][state2][state3];

    ll b = 1LL << dig;

    int d = ((N & b) != 0);

    mint res = 0;
    if (state1 == 0) {
        if (d == 0) res = f(dig - 1, 0, state2, state3); // 0 0
        else {
            if(state2 == 1) res += f(dig - 1, 0, 1, 1); // 1 1
            res += f(dig - 1, 1, 1, state3); // 0 1
            res += f(dig - 1, 2, state2, state3); // 0 0
        }
    } 
    else if (state1 == 1) { // 1:x<y=N
        if (d == 0) {
            res += f(dig - 1, 1, 1, state3); // 1 0
            res += f(dig - 1, 1, state2, state3); // 0 0
        }
        else {
            if (state2 == 1) res += f(dig - 1, 1, 1, 1); // 1 1
            res += f(dig - 1, 1, 1, state3); // 0 1
            res += f(dig - 1, 3, 1, state3); // 1 0
            res += f(dig - 1, 3, state2, state3); // 0 0
        }
    }
    else if (state1 == 2){ // 2:x=y<N
        if (state2 == 1) res += f(dig - 1, 2, 1, 1); // 1 1
        res += f(dig - 1, 3, 1, state3); // 0 1
        res += f(dig - 1, 2, state2, state3); // 0 0
    }
    else {
        if (state2 == 1) res += f(dig - 1, 3, 1, 1); // 1 1
        res += f(dig - 1, 3, 1, state3); // 0 1
        res += f(dig - 1, 3, 1, state3); // 1 0
        res += f(dig - 1, 3, state2, state3); // 0 0
    }

    vis[dig][state1][state2][state3] = 1;
    return memo[dig][state1][state2][state3] = res;
}
//---------------------------------------------------------------------------------------------------
void _main() {
    cin >> N;
    cout << f(60, 0, 0, 0) << endl;
}
0