結果
| 問題 | No.685 Logical Operations | 
| コンテスト | |
| ユーザー |  | 
| 提出日時 | 2018-05-11 22:59:34 | 
| 言語 | C++14 (gcc 13.3.0 + boost 1.87.0) | 
| 結果 | 
                                AC
                                 
                             | 
| 実行時間 | 2 ms / 2,000 ms | 
| コード長 | 3,732 bytes | 
| コンパイル時間 | 1,675 ms | 
| コンパイル使用メモリ | 174,336 KB | 
| 実行使用メモリ | 6,944 KB | 
| 最終ジャッジ日時 | 2024-06-28 08:57:11 | 
| 合計ジャッジ時間 | 2,606 ms | 
| ジャッジサーバーID (参考情報) | judge4 / judge5 | 
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| ファイルパターン | 結果 | 
|---|---|
| other | AC * 27 | 
ソースコード
#define _USE_MATH_DEFINES
#include "bits/stdc++.h"
using namespace std;
#define FOR(i,j,k) for(int (i)=(j);(i)<(int)(k);++(i))
#define rep(i,j) FOR(i,0,j)
#define each(x,y) for(auto &(x):(y))
#define mp make_pair
#define MT make_tuple
#define all(x) (x).begin(),(x).end()
#define debug(x) cout<<#x<<": "<<(x)<<endl
#define smax(x,y) (x)=max((x),(y))
#define smin(x,y) (x)=min((x),(y))
#define MEM(x,y) memset((x),(y),sizeof (x))
#define sz(x) (int)(x).size()
#define RT return
#define vv(a,b,c,d) vector<vector<a> >(b,vector<a>(c,d))
#define vvv(a,b,c,d,e) vector<vector<vector<a> > >(b,vv(a,c,d,e))
using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;
using vll = vector<ll>;
template<int MOD>
class ModInt {
public:
    ModInt() :value(0) {}
    ModInt(long long val) :value((int)(val<0 ? MOD + val%MOD : val%MOD)) { }
    ModInt& operator+=(ModInt that) {
        value = value + that.value;
        if (value >= MOD)value -= MOD;
        return *this;
    }
    ModInt& operator-=(ModInt that) {
        value -= that.value;
        if (value<0)value += MOD;
        return *this;
    }
    ModInt& operator*=(ModInt that) {
        value = (int)((long long)value * that.value % 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 pow(long long k) const {
        if (value == 0)return 0;
        ModInt n = *this, res = 1;
        while (k) {
            if (k & 1)res *= n;
            n *= n;
            k >>= 1;
        }
        return res;
    }
    ModInt inverse() const {
        long long a = value, b = MOD, u = 1, v = 0;
        while (b) {
            long long t = a / b;
            a -= t * b;
            swap(a, b);
            u -= t * v;
            swap(u, v);
        }
        return ModInt(u);
    }
    int toi() const { return value; }
private:
    int value;
};
typedef ModInt<1000000007> mint;
ostream& operator<<(ostream& os, const mint& x) {
    os << x.toi();
    return os;
}
mint dp[70][2][2][2][2][2];
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << fixed << setprecision(20);
    ll N;
    while (cin >> N) {
        if (N == 0) {
            cout << 0 << endl;
            continue;
        }
        vi A;
        ll p = N;
        while (p > 0) {
            A.push_back(p % 2);
            p /= 2;
        }
        reverse(all(A));
        int K = sz(A);
        dp[0][0][0][0][0][0] = 1;
        rep(i, K)rep(lx, 2)rep(ly, 2) rep(xy, 2)rep(land, 2)rep(lxor, 2) {
            rep(x, 2)rep(y, 2) {
                if (!lx && x > A[i])continue;
                if (!ly && y > A[i])continue;
                if (!xy && x > y)continue;
                int nlx = lx || x < A[i];
                int nly = ly || y < A[i];
                int nxy = xy || x < y;
                int pand = x & y;
                int pxor = x ^ y;
                int por = x | y;
                if (!land && pand > pxor)continue;
                if (!lxor && pxor > por)continue;
                int nland = land || pand < pxor;
                int nlxor = lxor || pxor < por;
                dp[i + 1][nlx][nly][nxy][nland][nlxor] += dp[i][lx][ly][xy][land][lxor];
            }
        }
        mint re;
        rep(lx, 2)rep(ly, 2)rep(xy, 2) {
            re += dp[K][lx][ly][xy][1][1];
        }
        cout << re << endl;
    }
}
            
            
            
        