1116.print-zero-even-odd

Statement

简体中文English

Metadata

• Link: 打印零与奇偶数
• Difficulty: Medium
• Tag: 多线程

现有函数 printNumber 可以用一个整数参数调用，并输出该整数到控制台。

• 例如，调用 printNumber(7) 将会输出 7 到控制台。

给你类 ZeroEvenOdd 的一个实例，该类中有三个函数：zero、even 和 odd 。ZeroEvenOdd 的相同实例将会传递给三个不同线程：

• 线程 A：调用 zero() ，只输出 0
• 线程 B：调用 even() ，只输出偶数
• 线程 C：调用 odd() ，只输出奇数

修改给出的类，以输出序列 "010203040506…" ，其中序列的长度必须为 2n 。

实现 ZeroEvenOdd 类：

• ZeroEvenOdd(int n) 用数字 n 初始化对象，表示需要输出的数。
• void zero(printNumber) 调用 printNumber 以输出一个 0 。
• void even(printNumber) 调用printNumber 以输出偶数。
• void odd(printNumber) 调用 printNumber 以输出奇数。

 

示例 1：

输入：n = 2
输出："0102"
解释：三条线程异步执行，其中一个调用 zero()，另一个线程调用 even()，最后一个线程调用odd()。正确的输出为 "0102"。

示例 2：

输入：n = 5
输出："0102030405"

 

提示：

• 1 <= n <= 1000

Metadata

• Link: Print Zero Even Odd
• Difficulty: Medium
• Tag: Concurrency

You have a function printNumber that can be called with an integer parameter and prints it to the console.

• For example, calling printNumber(7) prints 7 to the console.

You are given an instance of the class ZeroEvenOdd that has three functions: zero, even, and odd. The same instance of ZeroEvenOdd will be passed to three different threads:

• Thread A: calls zero() that should only output 0's.
• Thread B: calls even() that should only output even numbers.
• Thread C: calls odd() that should only output odd numbers.

Modify the given class to output the series "010203040506…" where the length of the series must be 2n.

Implement the ZeroEvenOdd class:

• ZeroEvenOdd(int n) Initializes the object with the number n that represents the numbers that should be printed.
• void zero(printNumber) Calls printNumber to output one zero.
• void even(printNumber) Calls printNumber to output one even number.
• void odd(printNumber) Calls printNumber to output one odd number.

 

Example 1:

Input: n = 2
Output: "0102"
Explanation: There are three threads being fired asynchronously.
One of them calls zero(), the other calls even(), and the last one calls odd().
"0102" is the correct output.

Example 2:

Input: n = 5
Output: "0102030405"

 

Constraints:

• 1 <= n <= 1000

Solution

C++

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define endl "\n"
#define fi first
#define se second
#define all(x) begin(x), end(x)
#define rall rbegin(a), rend(a)
#define bitcnt(x) (__builtin_popcountll(x))
#define complete_unique(a) a.erase(unique(begin(a), end(a)), end(a))
#define mst(x, a) memset(x, a, sizeof(x))
#define MP make_pair

using ll = long long;
using ull = unsigned long long;
using db = double;
using ld = long double;
using VLL = std::vector<ll>;
using VI = std::vector<int>;
using PII = std::pair<int, int>;
using PLL = std::pair<ll, ll>;

using namespace __gnu_pbds;
using namespace std;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

template <typename T, typename S>
inline bool chmax(T &a, const S &b) {
    return a < b ? a = b, 1 : 0;
}

template <typename T, typename S>
inline bool chmin(T &a, const S &b) {
    return a > b ? a = b, 1 : 0;
}

#ifdef LOCAL
#include <debug.hpp>
#else
#define dbg(...)
#endif
// head

class ZeroEvenOdd {
private:
    int n;

public:
    ZeroEvenOdd(int n) {
        this->n = n;
    }

    // printNumber(x) outputs "x", where x is an integer.
    void zero(function<void(int)> printNumber) {
        for (int i = 0; i < n; i++) {
            while (order_ % 2 != 0) {
                std::this_thread::yield();
            }

            printNumber(0);

            order_++;
        }
    }

    void even(function<void(int)> printNumber) {
        for (int i = 2; i <= n; i += 2) {
            while (order_ % 4 != 3) {
                std::this_thread::yield();
            }

            printNumber(i);

            order_++;
        }
    }

    void odd(function<void(int)> printNumber) {
        for (int i = 1; i <= n; i += 2) {
            while (order_ % 4 != 1) {
                std::this_thread::yield();
            }

            printNumber(i);

            order_++;
        }
    }

private:
    int order_{0};
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

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最后更新: October 11, 2023