結果

問題 No.2008 Super Worker
ユーザー 🍮かんプリン
提出日時 2022-07-15 23:45:58
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 148 ms / 2,000 ms
コード長 6,161 bytes
コンパイル時間 1,770 ms
コンパイル使用メモリ 174,328 KB
実行使用メモリ 6,528 KB
最終ジャッジ日時 2024-06-27 21:24:43
合計ジャッジ時間 4,703 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge2
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 33
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 #
プレゼンテーションモードにする

#pragma region include
#include <iostream>
#include <iomanip>
#include <stdio.h>
#include <vector>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <bitset>
#include <complex>
#include <deque>
#include <functional>
#include <limits>
#include <list>
#include <numeric>
#include <queue>
#include <stack>
#include <utility>
#include <array>
#include <random>
#include <tuple>
#include <unordered_map>
#include <unordered_set>
#include <assert.h>
#include <chrono>
#include <cstring>
#include <memory>
#include <unistd.h>
using namespace std;
typedef long long ll;
#define dump(x) cerr << #x << " = " << (x) << " [" << __LINE__ << ":" << __FUNCTION__ << "] " << endl
// vector???��?��??��?��o???��?��??��?��???��?��??��?��
template < typename T >
ostream& operator<<(ostream& os, vector< T >& v) {
    os << "{";
    for (int i = 0; i < (int)v.size(); i++) {
        os << v[i] << (i < v.size() - 1 ? ", " : "");
    }
    os << "}";
    return os;
}
// pair???��?��??��?��o???��?��??��?��???��?��??��?��
template < typename T, typename U >
ostream& operator<<(ostream& os, const pair< T, U >& p) {
    return os << "(" << p.first << ", " << p.second << ")";
}
// map???��?��??��?��o???��?��??��?��???��?��??��?��
template < typename T, typename U >
ostream& operator<<(ostream& os, const map< T, U >& map_var) {
    os << "{";
    for (auto itr = map_var.begin(); itr != map_var.end(); itr++) {
        os << "(" << itr->first << ", " << itr->second << ")";
        itr++;
        if (itr != map_var.end()) os << ", ";
        itr--;
    }
    os << "}";
    return os;
}
// set ???��?��??��?��o???��?��??��?��???��?��??��?��
template < typename T >
ostream& operator<<(ostream& os, const set< T >& set_var) {
    os << "{";
    for (auto itr = set_var.begin(); itr != set_var.end(); itr++) {
        os << *itr;
        ++itr;
        if (itr != set_var.end()) os << ", ";
        itr--;
    }
    os << "}";
    return os;
}
// multiset ???��?��??��?��o???��?��??��?��???��?��??��?��
template < typename T >
ostream& operator<<(ostream& os, const multiset< T >& set_var) {
    os << "{";
    for (auto itr = set_var.begin(); itr != set_var.end(); itr++) {
        os << *itr;
        ++itr;
        if (itr != set_var.end()) os << ", ";
        itr--;
    }
    os << "}";
    return os;
}
#pragma endregion
/////////////////////////////////////////
#include "bits/stdc++.h" 
using namespace std; 
typedef long long ll;
template< int MOD >
struct mint {
public:
    unsigned int x;
    mint() : x(0) {}
    mint(long long v) {
        long long w = (long long)(v % (long long)(MOD));
        if (w < 0) w += MOD;
        x = (unsigned int)(w);
    }
    mint(std::string &s) {
        unsigned int z = 0;
        for (int i = 0; i < s.size(); i++) {
            z *= 10;
            z += s[i] - '0';
            z %= MOD;
        }
        x = z;
    }
    mint operator+() const { return *this; }
    mint operator-() const { return mint() - *this; }
    mint& operator+=(const mint &a) {
        if ((x += a.x) >= MOD) x -= MOD;
        return *this;
    }
    mint& operator-=(const mint &a) {
        if ((x -= a.x) >= MOD) x += MOD;
        return *this;
    }
    mint& operator*=(const mint &a) {
        unsigned long long z = x;
        z *= a.x;
        x = (unsigned int)(z % MOD);
        return *this;
    }
    mint& operator/=(const mint &a) {return *this = *this * a.inv(); }
    friend mint operator+(const mint& lhs, const mint& rhs) {
        return mint(lhs) += rhs;
    }
    friend mint operator-(const mint& lhs, const mint& rhs) {
        return mint(lhs) -= rhs;
    }
    friend mint operator*(const mint& lhs, const mint& rhs) {
        return mint(lhs) *= rhs;
    }
    friend mint operator/(const mint& lhs, const mint& rhs) {
        return mint(lhs) /= rhs;
    }
    friend bool operator==(const mint& lhs, const mint& rhs) {
        return lhs.x == rhs.x;
    }
    friend bool operator!=(const mint& lhs, const mint& rhs) {
        return lhs.x != rhs.x;
    }
    friend std::ostream& operator<<(std::ostream &os, const mint &n) {
        return os << n.x;
    }
    friend std::istream &operator>>(std::istream &is, mint &n) {
        unsigned int x;
        is >> x;
        n = mint(x);
        return is;
    }
    mint inv() const {
        assert(x);
        return pow(MOD-2);
    }
    mint pow(long long n) const {        
        assert(0 <= n);
        mint p = *this, r = 1;
        while (n) {
            if (n & 1) r *= p;
            p *= p;
            n >>= 1;
        }
        return r;
    }
    
    // 存在しない場合0を返す(二乗して確認).
    // O(log^2MOD)
    mint sqrt() const {
        if (this->x < 2) return *this;
        if (this->pow((MOD-1)>>1).x != 1) return mint(0);
        mint b = 1, one = 1;
        while (b.pow((MOD-1) >> 1) == 1) b += one;
        long long m = MOD-1, e = 0;
        while (m % 2 == 0) m >>= 1, e += 1;
        mint x = this->pow((m - 1) >> 1);
        mint y = (*this) * x * x;
        x *= (*this);
        mint z = b.pow(m);
        while (y.x != 1) {
            int j = 0;
            mint t = y;
            while (t != one) j += 1, t *= t;
            z = z.pow(1LL << (e-j-1));
            x *= z; z *= z; y *= z; e = j;
        }
        return x;
    }
};
constexpr int MOD = 1e9 + 7;
int main() {
    using P = pair<ll,ll>;
    int n;cin >> n;
    vector<P> p(n);
    for (int i = 0; i < n; i++) {
        cin >> p[i].first;
    }
    for (int i = 0; i < n; i++) {
        cin >> p[i].second;
    }
    sort(p.begin(), p.end(), [](P a,P b){
        return (a.second-1)*b.first > (b.second-1)*a.first;
    });
    // dump(p);
    mint<MOD> ans = 0;
    mint<MOD> x = 1;
    for (int i = 0; i < n; i++) {
        ans += p[i].first * x;
        x *= p[i].second;
    }
    cout << ans << endl;
    return 0;
}
/////////////////////////////////////////
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