c++ - 当方程式返回 nan 作为答案时该怎么办？

Question

我的程序一直有一个小问题，我想做的是为用户开发一种方法来模拟可能的密码强度。这是假设所有密码都是排列(我知道这很奇怪，但我认为这是为了阻止数据变得更加笨拙。)使用等式......

//n!/(n-r)!当 n! = (e^-n)*(n^n) 平方根 (2(pi)n)。当n为使用的字符数，r为密码长度时

无论我输入什么，我都会收到 nan 作为答案。我认为也许我的方程式不对(也许我以某种方式除以零)所以我重新设计并大大简化了它。但这似乎不是问题所在，尽管我觉得这让我更接近于正确。但我想也许数字溢出在这里有影响？但我真的不知道如何解决这样的问题。我尝试从不同的数据类型跳转，但似乎没有任何效果。

我也有模数问题。它返回的时间小于零的数字，所以根据我的菜鸟知识告诉我，也许我又溢出了但是我还怎么使用 % 而不将它定义为 int？也许解决上述问题会解决这个问题？

如有任何帮助，我将不胜感激。如何处理 nan 的返回值？有没有一步步解决的现状？它几乎总是溢出还是可能是其他原因？

代码本身。

#include <iostream>
#include <cmath>

using namespace std;

const int SECONDS_IN_YEAR = 31556926;
const int SECONDS_IN_DAY  = 86400;
const int SECONDS_IN_HOUR = 3600;
const int SECONDS_IN_MIN  = 60;

int main()
{

    int passwordLength ,characterSymbols;
    double instructionsPerSecond, instructionSuccess;

    ////////////////////////////////////////////////////////////////////////////////
    //Equations needed
    // n!/(n-r)!
    //n is the number of letters in the alphabet
    //and r is the number of letters in the password

    // n! = (e^-n)*(n^n) sqrt(2(pi)n)
    double numeratorFactorial  = (pow(M_E,-characterSymbols))
                                *(pow(characterSymbols,characterSymbols))
                                *(sqrt(2*M_PI*characterSymbols));
    // (n-r)
    double characterMinusLength= (characterSymbols-passwordLength);

    // (n-r)! = (e^-(n-r)) * ((n-r)^(n-r)) * sqrt(2(pi)(n-r))
    double denominatorFactorial = ((pow(M_E, -(characterMinusLength)))*
                                (pow((characterMinusLength),(characterMinusLength)))
                                * (sqrt(2*M_PI*(characterMinusLength))));

    // n!/(n-r)!
    long double passwordPermutation = (numeratorFactorial / denominatorFactorial);

    // (passwords)* (instructions/Password) * (seconds/instruction) = sec
    int passwordSeconds = (passwordPermutation * instructionSuccess)
                         *(1/instructionsPerSecond);



    int passwordMin  = passwordSeconds / SECONDS_IN_MIN ;
    int passwordHour = passwordSeconds / SECONDS_IN_HOUR;
    int passwordDay  = passwordSeconds / SECONDS_IN_DAY ;
    int passwordYear = passwordSeconds / SECONDS_IN_YEAR;


    ////////////////////////////////////////////////////////////////////////////////
    //Explain purpose of program
    cout << "This program is designed to simulate the strength of passwords." << endl;

    //Ask for alphabet
    cout << "But first, share with me the max number of characters you'd be using."
         << endl;
    cin >> characterSymbols;
    //Reflect information
    cout << "We will be using " << characterSymbols << " character symbols to "
         << " construct the password.\n" << endl;



    ///////////////////////////////////////////////////////////////////////////////
    //Input length of password
    cout << "\n\nWill you give me the length of proposed password?" << endl;
    cin >> passwordLength;

    //Repeat information
    cout << "The password length will be " << passwordLength << "." <<endl;



    //cout permutations
    cout << "This would lead to " << passwordPermutation << " unique password\n"
         << endl;



    ////////////////////////////////////////////////////////////////////////////////
    //Ask for computer strength
    cout << "How powerful is this computer? How many instructions per second " << endl;
    cout << "can it accomplish?" << endl;
    cin >> instructionsPerSecond;

    //Read out computer strength
    cout << "The computer can do " << instructionsPerSecond << " instructions/second"
         << endl << endl;



    ////////////////////////////////////////////////////////////////////////////////

    //Ask for instructions/password
    cout << "The number of instructions needed to test your password is." << endl
         << endl;
    cin >> instructionSuccess;

    //reflect
    cout << "This computer can do " << instructionSuccess
     << " instructions/password" << endl;


    ////////////////////////////////////////////////////////////////////////////////
    cout << "\n\nThe amount of seconds it'll take to crack this passcode is... "
         << endl << passwordSeconds << " seconds.\n\n\n\n\n" << endl;

    ////////////////////////////////////////////////////////////////////////////////
    //Reflect all information in an easily readable table
    cout << "Number of character symbols using... " << characterSymbols      << endl;
    cout << "Length of password... "                << passwordLength        << endl;
    cout << "Number of permutations... "            << passwordPermutation   << endl;
    cout << "Instructions per second... "           << instructionsPerSecond << endl;
    cout << "Instructions per password..."          << instructionSuccess    << endl;

    cout << endl << endl << endl;

    ////////////////////////////////////////////////////////////////////////////////
    //Add in conversions for min, hour, day, years
    cout << "Number of seconds to break..." << passwordSeconds << endl;


    cout << "Converted to minutes..."       << passwordMin     << endl;
    passwordMin = passwordSeconds / SECONDS_IN_MIN;
    passwordSeconds = passwordSeconds % SECONDS_IN_MIN;

    cout << "Converted to hours..."         << passwordHour    << endl;
    passwordHour = passwordSeconds / SECONDS_IN_HOUR;
    passwordSeconds = passwordSeconds % SECONDS_IN_MIN;

    cout << "Converted to days..."          << passwordDay     << endl;
    passwordDay = passwordSeconds / SECONDS_IN_DAY;
    passwordSeconds = passwordSeconds % SECONDS_IN_DAY;

    cout << "Converted to years..."         << passwordYear    << endl;
    passwordYear = passwordSeconds / SECONDS_IN_YEAR;
    passwordSeconds = passwordSeconds % SECONDS_IN_YEAR;

    return (0);
}

Answer 1

“nan”代表“不是数字”。发生这种情况是因为您声明了变量 characterSymbols 和 passwordLength 而没有给它们初始值。

您必须在使用任何变量之前对其进行初始化 - 如果您不这样做，那么您将有未确定的行为。例如:

int x;
int y;
int z = x + y;

这里无法预测 z 等于什么，因为我们不知道 x 或 y 等于什么。同样，您的代码应该是这样的:

int characterSymbols = 10;  //or whatever you want the initial value to be

...

double numeratorFactorial  = (pow(M_E,-characterSymbols))
                            *(pow(characterSymbols,characterSymbols))
                            *(sqrt(2*M_PI*characterSymbols));

这样，numeratorFactorial 就会有一个有效的值。