一.算法类

1622题,困难,奇妙序列

class Fancy {
public:
    static const int MOD = 1e9 + 7;
    long long M_total; // cumulative multiplicative factor
    long long A_total; // cumulative additive factor
    vector<long long> val; // original values
    vector<long long> M_i; // multiplicative factor at time of append
    vector<long long> A_i; // additive factor at time of append
    
    Fancy() {
        M_total = 1;
        A_total = 0;
    }
    
    // Function to compute x^y % MOD
    long long modPow(long long x, long long y) {
        long long result = 1;
        x %= MOD;
        while (y > 0) {
            if (y % 2 == 1)
                result = result * x % MOD;
            x = x * x % MOD;
            y /= 2;
        }
        return result;
    }
    
    // Function to compute modular inverse of a modulo MOD
    long long modInverse(long long a) {
        return modPow(a, MOD - 2);
    }
    
    void append(int val_) {
        val.push_back(val_);
        M_i.push_back(M_total);
        A_i.push_back(A_total);
    }
    
    void addAll(int inc) {
        A_total = (A_total + inc) % MOD;
    }
    
    void multAll(int m) {
        M_total = (M_total * m) % MOD;
        A_total = (A_total * m) % MOD;
    }
    
    int getIndex(int idx) {
        if (idx >= val.size())
            return -1;
        long long v_i = val[idx];
        long long M_i_inv = modInverse(M_i[idx]);
        long long M = M_total * M_i_inv % MOD;
        long long A_i_mul_M = A_i[idx] * M % MOD;
        long long A = (A_total - A_i_mul_M + MOD) % MOD;
        long long result = (v_i * M % MOD + A) % MOD;
        return (int)result;
    }
};

通关截图 

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2043题,中等,简易银行系统

class Bank {
public:
    vector<long long> balance; // 存储每个账户的余额

    Bank(vector<long long>& balance_) {
        balance = balance_; // 初始化账户余额
    }
    
    bool transfer(int account1, int account2, long long money) {
        // 检查账户是否有效
        if (account1 < 1 || account1 > balance.size() || account2 < 1 || account2 > balance.size())
            return false;
        // 检查账户1的余额是否足够
        if (balance[account1 - 1] < money)
            return false;
        // 执行转账操作
        balance[account1 - 1] -= money;
        balance[account2 - 1] += money;
        return true;
    }
    
    bool deposit(int account, long long money) {
        // 检查账户是否有效
        if (account < 1 || account > balance.size())
            return false;
        // 执行存款操作
        balance[account - 1] += money;
        return true;
    }
    
    bool withdraw(int account, long long money) {
        // 检查账户是否有效
        if (account < 1 || account > balance.size())
            return false;
        // 检查账户余额是否足够
        if (balance[account - 1] < money)
            return false;
        // 执行取款操作
        balance[account - 1] -= money;
        return true;
    }
};

通关截图

 

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二.数据库

        262题,困难,行程和用户

SELECT
    t.request_at AS Day,
    ROUND(
        AVG(t.status IN ('cancelled_by_driver', 'cancelled_by_client')), 2
    ) AS 'Cancellation Rate'
FROM
    Trips t
JOIN Users cu ON t.client_id = cu.users_id AND cu.banned = 'No'
JOIN Users du ON t.driver_id = du.users_id AND du.banned = 'No'
WHERE
    t.request_at BETWEEN '2013-10-01' AND '2013-10-03'
GROUP BY
    t.request_at;

通关截图