The big wheel winning probability algorithm is often encountered in our daily life. So how to implement the winning probability algorithm based on PHP code? Here is a code example to introduce the PHP winning probability algorithm. The code is simple and easy to understand, and comes with comments. The specific code is as follows:
<?php /* * 经典的概率算法, * $proArr是一个预先设置的数组, * 假设数组为:array(100,200,300,400), * 开始是从1,1000 这个概率范围内筛选第一个数是否在他的出现概率范围之内, * 如果不在,则将概率空间,也就是k的值减去刚刚的那个数字的概率空间, * 在本例当中就是减去100,也就是说第二个数是在1,900这个范围内筛选的。 * 这样 筛选到最终,总会有一个数满足要求。 * 就相当于去一个箱子里摸东西, * 第一个不是,第二个不是,第三个还不是,那最后一个一定是。 * 这个算法简单,而且效率非常 高, * 关键是这个算法已在我们以前的项目中有应用,尤其是大数据量的项目中效率非常棒。 */ function get_rand($proArr) { $result = ''; //概率数组的总概率精度 $proSum = array_sum($proArr); //概率数组循环 foreach ($proArr as $key => $proCur) { $randNum = mt_rand(1, $proSum); if ($randNum <= $proCur) { $result = $key; break; } else { $proSum -= $proCur; } } unset ($proArr); return $result; } /* * 奖项数组 * 是一个二维数组,记录了所有本次抽奖的奖项信息, * 其中id表示中奖等级,prize表示奖品,v表示中奖概率。 * 注意其中的v必须为整数,你可以将对应的 奖项的v设置成0,即意味着该奖项抽中的几率是0, * 数组中v的总和(基数),基数越大越能体现概率的准确性。 * 本例中v的总和为100,那么平板电脑对应的 中奖概率就是1%, * 如果v的总和是10000,那中奖概率就是万分之一了。 * */ $prize_arr = array( '0' => array('id'=>1,'prize'=>'平板电脑','v'=>1), '1' => array('id'=>2,'prize'=>'数码相机','v'=>5), '2' => array('id'=>3,'prize'=>'音箱设备','v'=>10), '3' => array('id'=>4,'prize'=>'4G优盘','v'=>12), '4' => array('id'=>5,'prize'=>'10Q币','v'=>22), '5' => array('id'=>6,'prize'=>'下次没准就能中哦','v'=>50), ); /* * 每次前端页面的请求,PHP循环奖项设置数组, * 通过概率计算函数get_rand获取抽中的奖项id。 * 将中奖奖品保存在数组$res['yes']中, * 而剩下的未中奖的信息保存在$res['no']中, * 最后输出json个数数据给前端页面。 */ foreach ($prize_arr as $key => $val) { $arr[$val['id']] = $val['v']; } $rid = get_rand($arr); //根据概率获取奖项id $res['yes'] = $prize_arr[$rid-1]['prize']; //中奖项 unset($prize_arr[$rid-1]); //将中奖项从数组中剔除,剩下未中奖项 shuffle($prize_arr); //打乱数组顺序 for($i=0;$i<count($prize_arr);$i++){ $pr[] = $prize_arr[$i]['prize']; } $res['no'] = $pr; print_r($res);
Let me share with you an example code based on Java to calculate the winning probability
When doing mobile projects, there is a demand, so let’s do a lottery! The calculation of winnings is quite disgusting. Users have to change the winning probability of each award, and there is a limit on the number of awards per day. The probabilities of prizes one, two, three, four, five and six are unreasonable. How can one calculate whether a user has won or not? After thinking hard, I can use the nextInt(int x) method of the Random class to generate a random number within a range. The winning range will be the winning range. The winning range is generated dynamically. The source code is posted for reference only!
package Mzone; import java.util.ArrayList; import java.util.Random; public class Mzone { /** * CopyRright(c)2009-04: * Project: * Module ID: * Comments: 概率计算 * JDK version used: <JDK1.4> * Author:ch * Create Date:2009-04-20 * Modified By: * Modified Date: * Why & What is modified * Version: 1.0 */ static Random r = new Random(); public static void main(String[] args) { //各个奖项的中奖概率的分母 Integer _5m = new Integer(5); Integer _500m = new Integer(30); Integer _ipod = new Integer(500); Integer _phone = new Integer(1000); Integer _notebook = new Integer(1500); Integer _jay = new Integer(50); ArrayList list = new ArrayList(); if(_5m.intValue()!=0) list.add(_5m); if(_500m.intValue()!=0) list.add(_500m); if(_ipod.intValue()!=0) list.add(_ipod); if(_phone.intValue()!=0) list.add(_phone); if(_notebook.intValue()!=0) list.add(_notebook); if(_jay.intValue()!=0) list.add(_jay); //计算最小公倍数 int common = getN(list); System.out.println("最小公倍数:"+common); int a = 0;int b = 0;int c = 0;int d = 0;int e = 0;int f = 0;int g = 0; int first = 0;int second = 0;int third = 0;int four = 0;int fifth = 0;int sixth = 0; if(_5m.intValue()!=0){ first = common/_5m.intValue(); } if(_500m.intValue()!=0){ second = first + (common/_500m.intValue()); }else second = first; if(_ipod.intValue()!=0){ third = second + (common/_ipod.intValue()); }else third = second; if(_phone.intValue()!=0){ four = third + (common/_phone.intValue()); }else four = third; if(_notebook.intValue()!=0){ fifth = four + (common/_notebook.intValue()); }else fifth = four; if(_jay.intValue()!=0){ sixth = fifth + (common/_jay.intValue()); }else sixth = fifth; int times = 30000;//循环次数 for(int i = 0;i < times; i++){ int ri = getRandom(common);//产生随机数 if(ri >= 0 && ri < first){ a++; }else if(ri >= first && ri < second){ b++; }else if(ri >= second && ri < third){ c++; }else if(ri >= third && ri < four){ d++; }else if(ri >= four && ri < fifth){ e++; }else if(ri >= fifth && ri < sixth){ f++; }else{ g++; } } System.out.println("5m值:" + a + " 500m值:" + b + " ipodMP3:" + c + " 手机:" + d + " 笔记本电脑:" + e + " 演唱会门票:" + f + " 谢谢参与:" + g); } /** * 求最大公约数 */ public static int gcd(int m, int n){ while (true){ if ((m = m % n) == 0) return n; if ((n = n % m) == 0) return m; } } /** * 求最小公倍数 */ public static int gys(int z, int y){ int t = 0; int c = 0; c = gcd(z,y); t = z * y / c; return t; } /** * 求几个数的最小公倍数 */ public static int getN(ArrayList list){ int t = 1; for(int i = 0;i<list.size();i++){ Integer temp = (Integer)list.get(i); t = gys(t,temp.intValue()); } return t; } /** * 产生随机数 */ public static int getRandom(int y){ int result = r.nextInt(y); return result; } }