C++ 生日概率

Question

我正在尝试自学 C++，为今年秋天的研究生院做准备，但我在解决这个生日悖论问题时遇到了一些麻烦。我的代码似乎运行正常，但我没有得到正确的输出。如果有人有任何建议，请告诉我。

 #include <cstdlib>
 #include <iostream>
 #include <ctime>

 using namespace std;

int main()
{
    srand(time(NULL));

    const int trials = 100000;
    int birthdays[50];
    int numMatches;


    for(int i = 2; i <= 50; i++)
    {
        numMatches = 0;

        for(int j = 1; j <= trials; j++)
        {

            for(int k = 1; k <= i; k++)
            {
                birthdays[k] = (rand() % 365) + 1;
            }

            int m = 1;
            bool matched = false;
            while(m < i && !matched){
                int n = m + 1;

            while(n <= i && !matched){
                if(birthdays[m] == birthdays[n]){
                    numMatches++;
                    matched = true;
                }
                n++;
            }
                m++;
            }
        }

        cout << "Probability of " << i << " people in a room sharing a birthday is \t"
          << ( float(numMatches) / float(trials) ) << endl;
    }  
}

Answer 1

您的代码没有计算一个 50 人的房间里有两个人同一天生日的概率。有几个错误，主要是索引，但这是最大的问题:

for(int j = 1; j <= trials; j++) {
    // assigns a random birthday to the first i people (should be 0 indexed)
    for(k = 1; k <= i; k++)
        birthdays[k] = (rand() % 365) + 1;
    // Does *exactly* the same thing as the previous loop, overwriting what
    // the initial loop did. Useless code
    for(m = 1; m <= i; m++)
        birthdays[m] = (rand() % 365) + 1;
    // At this point, m = k = i + 1. Here you check if
    // the i + 1st array value has the same b-day. It will, because they're
    // the same thing. Note you never set the i + 1st value so the loops
    // above did nothing
    if(birthdays[k] == birthdays[m])
        ++numMatches;
}

所以你在这里得到的是这样的:

• 执行以下 48 次迭代(从第一个循环开始，从 2 到 50:不知道这些值从何而来)
  • 对于这 48 次迭代中的每一次，执行 10k 次迭代:
    • 将一堆随机的东西分配给一个覆盖东西的数组
    • 忽略您在数组中写入的值，进行始终为真的比较并将 numMatches 递增 1