将倒数计时器添加到数学程序测验中
Add a countdown timer to a math program quiz
我正在尝试在此程序中添加倒数计时器。我希望计时器在提出第一个数学事实问题时开始,并且在到期后,我希望该程序给出等级。如果可能的话,在C 中执行什么代码?
#include <iostream>
#include <cstdlib>
#include <ctime>
#include <cstring>
using namespace std;
int main(int args, char* argv[])
{
int i;
int result;
int solution;
char fact;
bool done = false;
int correct = 0;
int count = 0;
do {
try {
cout << "Enter (m)ultiplication or "
<< "(a)ddition." << endl; /*or (s)ubstraction. */
cin >> fact;
while (!cin)
throw fact;
if (fact != 'A')
if (fact != 'a')
if (fact != 'M')
if (fact != 'm')
while (!cin)
throw fact;
cout << "Now, enter the number of the fact that
you would like to do." << endl;
cin >> i;
int wrong = 0;
int score = 0;
int j = 0;
while (!cin)
throw i;
switch (fact) {
case 'm':
case 'M':
while (j < 13) {
cout << "What's " << i << " x " << j << "?" << endl;
cin >> result;
while (!cin)
throw result;
solution = i * j;
if (result == solution) {
cout << "Great Job! That is the correct answer for the problem "
<< i << " x " << j << "." << endl;
cout << endl;
cout << endl;
cout << endl;
score++;
j++;
cout << endl;
}
if (result != solution) {
cout << "Oh no! " << result << " is NOT the correct answer for "
<< i << " x " << j << "." << endl;
wrong = wrong + 1;
count++;
}
if (count == 3) {
cout << "The correct answer is " << i * j << "." << endl;
j++;
wrong = wrong - 3;
count = 0;
}
if (count == 1) {
cout << endl;
count--;
wrong = wrong - 1;
}
if (count == 2) {
cout << endl;
count--;
wrong = wrong - 2;
}
}
case 'a':
case 'A':
while (j < 13) {
cout << "What's " << i << " + " << j << "?" << endl;
cin >> result;
while (!cin)
throw result;
solution = i + j;
if (result == solution) {
cout << "Great Job! That is the correct answer for the problem "
<< i << " + " << j << "." << endl;
cout << endl;
cout << endl;
cout << endl;
score++;
j++;
cout << endl;
}
if (result != solution) {
cout << "Oh no! " << result << " is NOT the correct answer for "
<< i << " + " << j << "." << endl;
wrong = wrong + 1;
count++;
}
if (count == 3) {
cout << "The correct answer is " << i + j << "." << endl;
j++;
wrong = wrong - 3;
count = 0;
}
if (count == 1) {
cout << endl;
count--;
wrong = wrong - 1;
}
if (count == 2) {
cout << endl;
count--;
wrong = wrong - 2;
}
}
if (j == 13) {
system("pause");
correct = score - wrong;
score = (correct * 100) / 13;
}
if (score >= 80) {
cout << "Excellent!!!!!" << endl;
cout << "You scored " << score << "%." << endl;
cout << "You got " << correct << " out of 13 correct." << endl;
cout << "Keep up the good work." << endl;
} else if (score >= 70) {
cout << "Congratulations!!!!!" << endl
cout << "You scored " << score << "%." << endl;
cout << "You got " << correct << " out of 13 correct." << endl;
cout << "Let's see if we can score even higher next time." << endl;
} else {
cout << "You scored below 70 which means that you may need some"
<< " more practice." << endl;
cout << "You scored " << score << "%." << endl;
cout << "You got " << correct << " out of 13 correct." << endl;
cout << "You might want to try the " << i << " facts again."
<< " Goodluck!!!!!" << endl;
}
}
} catch (char fact) {
cout << "Invalid input. You can only enter (m)ultiplication or"
<< " (a)ddition. Please try again." << endl;
cin.clear();
cin.ignore(100, 'n');
} catch (int i) {
cout << "Invalid input0. You can only enter a
number here. Please try again." << endl;
cin.clear();
cin.ignore(100, 'n');
} catch (...) {
cout << "Invalid input2. You can only enter a number here.
Please try again." << endl;
cin.clear();
cin.ignore(100, 'n');
}
} while (!done);
return 0;
}
任务很难,但是如果您敢尝试,我建议以两个步骤进行操作:
实施不准确的解决方案:对用户的查询之间检查计时器到期。
如果剩下一些时间,则提出下一个问题,否则会显示统计信息。因此,尽管计时器用完了,但程序始终在最后一个问题上等待用户输入。不是测验的样子,而是从。
开始的好举动方法:在开始测验保存当前时间之前,在每个问题之前,请在保存的时间和当前一个问题之间进行三角洲,并与时间限制进行比较。用Chrono(从C 11开始)的示例,以Oldschool Clock
为例添加中等问题中断
此部分需要功能,该功能将等待用户输入不长,而不是指定的时间。因此,您不必使用std :: cin()您需要计算剩余时间(时间限制在cur时间和开始时间之间的时间限制)并致电某种cin_with_timeout(time_left)。
最困难的是实现CIN_WITH_TIMEOUT(),它需要多线程和线程同步的可靠知识。可以在这里找到伟大的灵感,但这是开始思考而不是完整解决方案的方向。
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