c++生日悖论程序

C++ Birthday Paradox Program

本文关键字：程序 生日 c++      更新时间：2023-10-16

我正试图为我的c++类找出生日悖论程序。这是我目前的方法，但行不通。我试着在这里寻找关于同一主题的其他问题，但我仍然很迷路，所以任何帮助都会非常感激。

//This program calculates the likelihood of any two people in groups of 2-50 people 
//having their birthday on the same day, also known as the Birthday Paradox.
#include <iostream>
using namespace std;
int main()
{
int people, trial, count = 0, birthdays[50], numMatches, NUM_TRIALS = 5000;
double total;
//function call
sameBirthday(birthdays, people);
numMatches = 0;
for (people = 2; people <= 50; people++)
{
    //Run trials to see if people have the same birthday
    //Reset number of matches
    numMatches = 0;
    for (trial = 0; trial < NUM_TRIALS; trial++)
    {
        //Randomly generate up to "people" birthdays
        for (int i = 0; i < people, i++)
        {
            birthdays[people] = (rand() % 365) + 1;
            //Check to see if any two people have the same birthday
            for (i = 1; i < people; i++)
            {
                //birthday one
                for (int j = 0; j < i-1; j++)
                {
                    //birthday two
                    for (int k = j +1; k < i; k++);
                }
            }
        }
    }
}
bool sameBirthday(int birthdays[], int people)
{
    //if the two birthdays are the same, add one to the count 
    if (birthdays[j] == birthdays[k])
    {
        people++;
    }
    total = (numMatches / 5000.0);
    cout << "For " << people << " people, the probability of two birthdays is about " << total << endl;
}
return 0;

for (people = 2; people <= 50; people++)
{
    //Run trials to see if people have the same birthday
    //Reset number of matches
    numMatches = 0;
    for (trial = 0; trial < NUM_TRIALS; trial++)
    {
        //Randomly generate up to "people" birthdays
        for (int i = 0; i < people, i++)
        {
            //here I use i instead of people so every time i put the new number in a different position
            birthdays[i] = (rand() % 365) + 1;
            //this loop check if some birthday is equal to the one just generated
            for(int j = 0; j < i; j++){
                if(birthday[j] == birthday[i]){
                    //here do what u want to do when 2 people have the same birthday
                }
            }                
        }
    }
}

尝试使用这个循环，通过这种方式，如果有两个人生日相同，则检查从2到50的每个数字。

相关文章：