CLS打击表现

问题描述 投票:-1回答:2

此代码打印所需的no(4 here)完美数字。这是我的旧代码,检查每个数字是否完美。它运行正常并在6秒内完成,但是如果我在检查每个数字之前放置一个cls,它会受到很大的打击(大约100秒)。请告诉我背后的理论(为什么多重cls会影响性能)。

没有normie的答案就像“Dude显然需要点击,每次都刷新屏幕。”

PS-我知道更好的方法不是检查完美数字而是生成它们。(欧几里德 - 欧拉定理)

IDE-DevC ++

#include <iostream>
#include <string>
using namespace std;

int choice, quan, i, a[50], number;
string schoice;

void startgame()
{
    choice = 3;
    quan = 4;
    system("cls");
    cout << "Hey guys, today we are gonna print different types of numbers\n";
    cout << "\nFollowing are some of the special numbers found in the known observable little universe of ours:- \n";
    cout << "\n1. Prime numbers\n2. Square numbers\n3. Perfect Numbers\n\n";
    cout << "Which ones do you wanna see?\n";
    //cin >> choice;
    cout << "\nCool! How many of them do you wanna see?\n";
    //cin >> quan;
}
void perfect()
{
    if (choice == 3)
    {
        int j = 0, y = 0, f = 0, number = 2;
        do
        {
            //This is the cls in question
            //system("cls");
            cout << "The number under inspection: " << number << "\n";
            f = 0;
            for (i = (number - 1); i >= 1; i--)
            {
                if (number % i == 0)
                {
                    f = f + i;
                }
            }

            if (f == number)
            {
                //cout<<number<<", ";
                a[j] = number;
                j = j + 1;
                y = y + 1;
            }
            number++;
        }
        while (y < quan);
        system("cls");
        cout << "\nHere are your " << quan << " perfect numbers starting from 1:-\n";
        j = 1;
        for (i = 0; i < quan; i++)
        {
            cout << a[i];
            if (j != quan)
            {
                cout << ", ";
            }
            j++;
        }
    }
}

int main()
{
    do
    {
        startgame();
        perfect();
        cout << "\n\nReturn to Main menu?(Y/N)\n";
        schoice = "N";
        //cin >> schoice;
    }
    while (schoice == "Y");
    return 0;
}
c++ time-complexity
2个回答
0
投票

当您运行system(“cls”)时,您将创建一个新进程,并且每次选择一个新数字进行检查时都会这样做。创建一个进程是一项昂贵的操作,更重要的是你不需要它,如果你想要的只是更新屏幕上的当前数字。写作就足够了

cout << "The number under inspection: " << number << "\r";

cout.flush();

第一行将输出文本并将光标返回到同一行的开头。第二行将确保显示所有上述内容。注意:这也是运行缓慢,因此您可能希望对其进行速率限制。我建议你输出1000左右的每1个数字。


0
投票

Windows CreateProcess()相当昂贵,比Unix fork()贵得多。而且,系统(“cls”)是一种低效,不可移植的黑客。您不能在非Windows系统中使用它。您可以使用支持控制台操作的ncurses库。

对于Windows,通过Console API:

void clear() {
    COORD topLeft  = { 0, 0 };
    HANDLE console = GetStdHandle(STD_OUTPUT_HANDLE);
    CONSOLE_SCREEN_BUFFER_INFO screen;
    DWORD written;

    GetConsoleScreenBufferInfo(console, &screen);
    FillConsoleOutputCharacterA(
        console, ' ', screen.dwSize.X * screen.dwSize.Y, topLeft, &written
    );
    FillConsoleOutputAttribute(
        console, FOREGROUND_GREEN | FOREGROUND_RED | FOREGROUND_BLUE,
        screen.dwSize.X * screen.dwSize.Y, topLeft, &written
    );
    SetConsoleCursorPosition(console, topLeft);
}
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