C++ DeathPrediction -第一个程序如何在for循环中添加precentage

Question

我决定了我的第一个C++项目，我会做一个死亡预测器，它将有准确的死亡机会，并且会给你一个生日，在这个生日中，你会一直每隔360毫秒得到一次生日，直到你最终死去。

我面临的问题是，我不知道如何将前倾图表纳入我的工作。我只添加了19条if语句中的1条，因为线程上的600行代码对nagivate来说是一种痛苦。我如何结合这些前倾图表，使程序工作？

//GOOOD LUCK
#include "stdafx.h"
#include <iostream>
#include <iomanip>
#include <cmath>
#include <string>
#include <Windows.h>
#include <stdlib.h>
#include <time.h>
#include <random>

using namespace std;

int main()
{
    random_device rd;
    mt19937 gen(rd());

    uniform_int_distribution<> dis1 (1,28440);         // Died under 1 year of
    //age.  1.161756638 % chance

    uniform_int_distribution<> dis2 (28441,33196);     // Died between 1-4 years of
    //age.  0.194279696 % chance

    uniform_int_distribution<> dis3 (33197,36033);     // Died between 5-9 years of
    //age.  0.115889718 % chance

    uniform_int_distribution<> dis4 (36034,39798);     // Died between 10-14 years of
    //age.  0.153797951 % chance

    uniform_int_distribution<> dis5 (39799,53501);     // Died between 15-19 years of
    //age.  0.559759184 % chance

    uniform_int_distribution<> dis6 (53502,74032);     // Died between 20-24 years of
    //age.  0.838678816 % chance

    uniform_int_distribution<> dis7 (74033,93600);     // Died between 25-29 years of
    //age.  0.799340854 % chance

    uniform_int_distribution<> dis8 (93601,115957);    // Died between 30-34 years of
    //age.  0.913264801 % chance

    uniform_int_distribution<> dis9 (115958,147377);   // Died between 35-39 years of
    //age.  1.283487819 % chance

    uniform_int_distribution<> dis10(147378,200742);   // Died between 40-44 years of
    //age.  2.179927672 % chance

    uniform_int_distribution<> dis11(200743,280125);   // Died between 45-49 years of
    //age.  3.242747089 % chance

    uniform_int_distribution<> dis12(280126,384272);   // Died between 50-54 years of
    //age.  4.254341371 % chance

    uniform_int_distribution<> dis13(384273,511750);   // Died between 55-59 years of
    //age.  5.207398478 % chance

    uniform_int_distribution<> dis14(511751,659573);   // Died between 60-64 years of
    //age.  6.038474388 % chance

    uniform_int_distribution<> dis15(659574,831809);   // Died between 65-69 years of
    //age.  7.035735454 % chance

    uniform_int_distribution<> dis16(831810,1057928);  // Died between 70-74 years of
    //age.  9.236823110 % chance

    uniform_int_distribution<> dis17(1057929,1365816); // Died between 75-79 years of
    //age. 12.57703684 % chance

    uniform_int_distribution<> dis18(1365817,1744593); // Died between 80-84 years of
    //age. 15.47280922 % chance

    uniform_int_distribution<> dis19(1744594,2447762); // Died at age 85 and above.
    //     28.72402438 % chance


    //REFERENCE: http://www.disastercenter.com/cdc/Death%20rates%202005.html


    for (double count = 0; count < 101; count++)
    {
        int var1 = dis1(gen);
        int var2 = dis2(gen);
        int var3 = dis3(gen);
        int var4 = dis4(gen);
        int var5 = dis5(gen);
        int var6 = dis6(gen);
        int var7 = dis7(gen);
        int var8 = dis8(gen);
        int var9 = dis9(gen);
        int var10 = dis10(gen);
        int var11 = dis11(gen);
        int var12 = dis12(gen);
        int var13 = dis13(gen);
        int var14 = dis14(gen);
        int var15 = dis15(gen);
        int var16 = dis16(gen);
        int var17 = dis17(gen);
        int var18 = dis18(gen);
        int var19 = dis19(gen);

        if (count < 102)
        {
            Sleep(360);
            cout << "\n\t\t    It's your birthday! You turned: " << count;

        }
        if (var1 > 1 && var1 > 28440)
        {
            cout << "\n\n\n\t\t\tYou died! Better luck next time!";
            cout << "\n\n     ";
            Sleep(1000);
            cout << "T   "; Sleep(100);
            cout << "H   "; Sleep(100);
            cout << "A   "; Sleep(100);
            cout << "N   "; Sleep(100);
            cout << "K   "; Sleep(100);
            cout << "S   "; Sleep(100);
            cout << "    "; Sleep(100);
            cout << "F   "; Sleep(100);
            cout << "O   "; Sleep(100);
            cout << "R   "; Sleep(100);
            cout << "    "; Sleep(100);
            cout << "P   "; Sleep(100);
            cout << "L   "; Sleep(100);
            cout << "A   "; Sleep(100);
            cout << "Y   "; Sleep(100);
            cout << "I   "; Sleep(100);
            cout << "N   "; Sleep(100);
            cout << "G   "; Sleep(100);
            cout << "    "; Sleep(100);
            cout << "\n\n\n\t\t\t  ";
            return 0;

Answer 1

你可以尝试这样的方法：

std::random_device rd;
std::mt19937 gen(rd());
std::uniform_real_distribution<double> distribution(0, 100.);

unsigned int age = 0;
for (;;) {
    auto percent = distribution(gen);
    if (percent <= get_die_probability(age)) {
        break;
    }
    ++age;
    std::cout << "It's your birthday! You turned: " << age << std::endl;
}
std::cout << "You die at " << age << std::endl;

现场演示

Answer 2

您需要一个统一分布对象的向量，而不是由数以百万计的单独命名的变量组成。

typedef uniform_int_distribution<> dt;
vector<dt> dis;

int low= 0;
for (...stuff...) {
  ++low;
  high= ...;
  dis.push_back(dt(low,high));
  }