C ++生日悖论程序[重复]

Posted 2023-03-23

【中文标题】C ++生日悖论程序[重复]

【英文标题】：C++ Birthday Paradox Program [duplicate]

【发布时间】：2015-09-01 19:12:40

【问题描述】：

我正在尝试为我的 C++ 课程找出生日悖论程序。这是我到目前为止所拥有的，但它不起作用。我尝试在这里查看有关同一主题的其他问题，但我仍然很迷茫，因此将不胜感激。

//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;

【问题讨论】：

可能重复（选择一个）"*** c++ birthday paradox"

当您使用调试器时，哪些语句存在问题？

这是一期：birthdays[people] =。变量people 在循环中没有变化，并且超出了数组的范围。您的意思是使用第一个 i 变量吗？

您有两个循环使用索引变量i。虽然编译器不糊涂，你呢？也许您应该更改i 变量之一的名称。

你的k 循环没有做任何事情，去掉它或者给它一些内容。

【参考方案1】：

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，这样，如果有 2 个人的生日相同，则检查从 2 到 50 的每个人数。