c++ - 当我输入一个数字时，无论我使用什么数字，我都会得到 1 。我怎样才能解决这个问题？

Question

//"This program will ask for the number of people in a group and then output percentage likelyhood that two birthdays occur on the same day." << endl << endl;


#include <iostream>
#include <cmath>
#include <cstdlib>
#include <ctime>
using namespace std;

int FindThePerctange(int [], int);
void RandArray(int, int []);
void Counter(int []);

int main (){
    int GroupNumber;
    int DayOfBirth [365] = {};
    char Percant;
    Percant = '%';
    cout << "This program will ask for the number of people in a group and then output percentage likelyhood that two birthdays occur on the same day." << endl << endl;
    cout << "How many in group (0 quits)? "; 
    cin >> GroupNumber;
    if (GroupNumber == 0){
        cout << "\nThanks for using this program.";
    srand(time(0));
    }
    else{
        cout << "In a group of " << GroupNumber << " the chances for two birthdays the same is " << FindThePerctange(DayOfBirth, GroupNumber) << Percant << "." << endl << endl;
        cout << "How many in group (0 quits)? "; 
        cin >> GroupNumber;
        cout << "In a group of " << GroupNumber << " the chances for two birthdays the same is " << FindThePerctange(DayOfBirth, GroupNumber) << Percant << "." << endl << endl;
        cout << "How many in group (0 quits)? " << endl;
        cin >> GroupNumber;
        cout << "Thanks for using this program.";
    }
    cin.get();
    cin.get();
    return 0;
}


int FindThePerctange(int DayOfBirth [], int GroupNumber){
    double Overlap = 0, Percentage;
    for (int d =0; d<=10000; d++){
        (GroupNumber, DayOfBirth);
        while (d <= 365){
            int j;
            j = 0;
            j++;
            if(DayOfBirth[j] >= 2){
                Overlap = Overlap + 1;
                j=365;
            }
        return j;
        }
    Percentage = (Overlap/10000)*100;
    return Percentage;
    }
}

void RandArray(int GroupNumber,  int DayOfBirth[]){
    int Day, d;
    while (d < GroupNumber){
        Day = rand()%365;
        DayOfBirth[Day] +=1;
        d++;
    }
}

void Counter(int DayOfBirth[]){
    for (int d = 0; d<=365; d++){
        DayOfBirth[d] = 0;
    }
}

Answer 1

在您的 FindThePerctange 方法中，您遇到了问题。在 while 循环中，声明变量 j，将其设置为 0，将其递增为 1，然后立即返回。这意味着你每次都会得到 1。

我认为您要做的是在 while 循环中每次递增 j。如果是，请将 j 的声明放在 while 循环之外，如下所示:

int FindThePerctange(int DayOfBirth [], int GroupNumber){
    double Overlap = 0, Percentage;
    int j;
    for (int d =0; d<=10000; d++){
        //(GroupNumber, DayOfBirth);
        j = 0;
        while (d <= 365){
            j++;
            if(DayOfBirth[j] >= 2){
                Overlap++;
                j=365;
            }
        }
    Percentage = (Overlap/10000)*100;
    return Percentage;
    }
}

这仍然不能使你的代码工作，但它至少不会返回 1。我不会为你重写你的整个程序(其他人可能会)但它不会做你想要的期待它做。

编辑:我想为您提供更多帮助，但您的代码存在内部缺陷。您最大的问题可能是您没有为 DayOfBirth 数组分配任何值。我建议返回并重新编码。你问的问题很简单。请考虑以下事项:

假设每个人的生日是 1 到 365 之间的随机数。那么第一个人的生日被指定为 A。第二个人有 365 分之一的机会碰上同一天。因此对于组大小 2，机会是 1/365，或大约 .27。假设小组人数为 3，并且第二个人没有击中第一个人，则第三个人有 2/365 的几率，或者大约 0.54% 的几率。假设随机数生成器是完全随机的，那么两个生日发生冲突的几率将是 (groupsize - 1)/365。

因此您的代码只需要接受 GroupSize。它被缩短为:

float FindThePerctange(int GroupNumber){
    return (GroupNumber - 1) / 365 * 100;
}