#533774 (C++17) No.737 PopCount

• 問題: No.737 PopCount
• ユーザー: SHIJOU
• 提出日時: 2020-08-20 21:22:23
• 言語: C++17 (gcc 13.3.0 + boost 1.87.0)
• 結果: AC
• 実行時間: 3 ms / 1,000 ms
• コード長: 2,471 bytes
• コンパイル時間: 2,153 ms
• コンパイル使用メモリ: 194,536 KB
• 最終ジャッジ日時: 2025-01-13 04:31:37
• ジャッジサーバーID （参考情報）: judge5 / judge2

ファイルパターン	結果
sample	AC * 3
other	AC * 15

ソースコード

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//#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
#define rep(i, n) for(int i=0; i<n; ++i)
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
using namespace std;
using ll = int64_t;
using ld = long double;
using P = pair<int, int>;
using vs = vector<string>;
using vi = vector<int>;
using vvi = vector<vi>;
template<class T> using PQ = priority_queue<T>;
template<class T> using PQG = priority_queue<T, vector<T>, greater<T> >;
const int INF = 0xccccccc;
const ll LINF = 922337203685477580LL;
template<typename T1, typename T2>
inline bool chmax(T1 &a, T2 b) {return a < b && (a = b, true);}
template<typename T1, typename T2>
inline bool chmin(T1 &a, T2 b) {return a > b && (a = b, true);}
template<typename T1, typename T2>
istream &operator>>(istream &is, pair<T1, T2> &p) { return is >> p.first >> p.second;}
template<typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) { return os << p.first << ' ' << p.second;}
const int mod = 1000000007;
//const int mod = 998244353;
struct mint {
    int64_t x;
    mint(int64_t x=0):x((x%mod+mod)%mod){}
    mint operator-() const { return mint(-x);}
    mint& operator+=(const mint a) {
        if ((x += a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator-=(const mint a) {
        if ((x += mod-a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
    mint operator+(const mint a) const { return mint(*this) += a;}
    mint operator-(const mint a) const { return mint(*this) -= a;}
    mint operator*(const mint a) const { return mint(*this) *= a;}
    mint pow(int64_t t) const {
        if (!t) return 1;
        mint a = pow(t>>1);
        a *= a;
        if (t&1) a *= *this;
        return a;
    }
    //for prime mod
    mint inv() const { return pow(mod-2);}
    mint& operator/=(const mint a) { return *this *= a.inv();}
    mint operator/(const mint a) {return mint(*this) /= a;}
};
istream& operator>>(istream& is, mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}
//head
ll n, nn;
mint ans, sum, cnt;
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n;
    nn = n;
    for(int i = 62; i >= 0; i--) {
        ans += sum*(1LL<<(i+1))*(1LL<<i) + cnt*((1LL<<i)*3-1)*(1LL<<i)/2;
        sum *= 4;
        sum += cnt;
        cnt *= 2;
        if(n>>i&1) {
            sum += (n-nn)>>i;
            nn -= 1LL<<i;
            ans += mint(nn+1)*(n-nn) + mint(nn+1)*nn/2;
            cnt += 1;
        }
        //cout << i << ' ' << ans << '_' << sum << ' ' << cnt << '\n';
    }
    cout << ans << '\n';
}
 
 
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