明日方舟的抽奖系统的c++实现，一开始想的随机数实现，就是每一段数的范围代表不同的星级，其中概率分为40%，50%，8%，2%，首先随机数如下，取0~999。
a=rank()%1000;
随后就是十抽保底机制了，对于这种，鄙人是先bool一个变量看前十抽保底是否生效，如果生效了那就 返回1，抽到六星，五星均返回值1。
一旦抽卡次数达到了10，且还未有六星，五星出现，那么在四星，三星的判断下，加以个返回五星的语句，据如同这样。
else if(a>99+o&&a<600+o) //这是四星判断的地方。
{
if(ten==9&&shichou==0) //这里是判断是否到十抽且前是否有六星或者五星
{
k++; //这里是统计五星个数。
shichou=1;//保底生效
wu(0); //进入五星抽奖
wushi++;//五十抽保底机制，如果没有六星就自增。
}
else //前面的条件未成立
{
si(0); //进入四星抽奖
r++; //统计四星
wushi++; //五十抽保底机制
ten++; //十抽保底机制
}这样十抽保底就实现了，后面则是五十抽开始的保底机制。五十抽以后未抽到6星则概率增加，所以鄙人定义了一个变量，让它记录6星保底是否生效，如果生效，那它就变化。
代码如下：
if(a>=0&&a<(20+o)) //这里是判断是否6星
{
liu(0); //进入六星抽奖
j++; //六星统计
if(shichou==0)shichou=1; //前十抽保底机制，有了六星就失效了。
wushi=0; //五十抽保底归O
o=0; //五十抽保底增加概率归O
}
当未抽到六星时候，且五十抽保底（变量wushi>50）则下面公式：
if(wushi>50)o=(wushi-50)*20;
使五十抽保底开始。
这样到了九十九抽未有6星（黑到极致），一定百抽到达你身边（这样好像也是反向百抽的欧皇呢）。
而通过随机数，也可以实现一些干员UP的环境。这里就不多阐述了。
直接上所有代码！
#include<bits/stdc++.h>
using namespace std;
int ten;
int wushi;
int o;
int a,j,k,r,h,huan,n;
bool shichou;
void san(int x)
{
x=rand()%16; //三星干员
string l[16]={"芬","香草","翎羽","玫兰莎","卡缇","米格鲁","克洛斯","炎熔","芙蓉","安塞尔","史都华德","梓兰","空爆",
"月见夜","斑点","泡普卡"};
if(huan==0){
cout<<l[x];
printf("[★★★]");
huan++;
}
else if(huan==4)
{
cout<<' '<<l[x];
printf("[★★★]\n");
huan=0;
}
else
{
cout<<' '<<l[x];
printf("[★★★]");
huan++;
}
return;
}
void si(int x)
{
x=rand()%41; //四星干员
string l[41]={"夜烟","远山","杰西卡","流星","白雪","清道夫","红豆","杜宾","缠丸","霜叶","慕斯","砾","暗索","末药",
"调香师","角峰","蛇尾箱","古米","深海色","地灵","阿消","猎蜂","格雷伊","苏苏洛","桃金娘","红云","梅","安比尔",
"宴","刻刀","波登可","卡达","孑","酸糖","芳汀","泡泡","杰克","松果","豆苗","深靛","罗比菈塔"};
if(huan==0){
cout<<l[x];
printf("[★★★★]");
huan++;
}
else if(huan==4)
{
cout<<' '<<l[x];
printf("[★★★★]\n");
huan=0;
}
else
{
cout<<' '<<l[x];
printf("[★★★★]");
huan++;
}
return;
}
void wu(int x)
{
x=rand()%60; //五星干员
string l[60]={"白面鸮","凛冬","德克萨斯","芙兰卡","拉普兰德","幽灵鲨","蓝毒","白金","陨星","天火","梅尔","赫默",
"华法琳","临光","红","雷蛇","可颂","普罗旺斯","守林人","崖心","初雪","真理","空","狮蝎","食铁兽","夜魔","诗怀雅",
"格劳克斯","星极","送葬人","槐琥","苇草","布洛卡","灰喉","吽","惊蛰","摄砂","巫恋","极境","石棉","月禾","莱因哈特",
"断崖","蜜蜡","贾维","安哲拉","燧石","四月","奥斯塔","絮雨","卡夫卡","爱丽丝","乌有","熔泉","赤东","绮良","羽毛笔",
"桑葚","灰毫","浊清"};
if(huan==0){
cout<<l[x];
printf("[★★★★★]");
huan++;
}
else if(huan==4)
{
cout<<' '<<l[x];
printf("[★★★★★]\n");
huan=0;
}
else
{cout<<' '<<l[x];
printf("[★★★★★]");
huan++;
}
return;
}
void liu(int x)
{
x=rand()%38; //六星干员
string l[38]={"能天使","推进之王","伊芙利特","艾雅法拉","安洁莉娜","闪灵","夜莺","星熊","塞雷娅","银灰",
"斯卡蒂","陈","黑","赫拉格","麦哲伦","莫斯提马","煌","阿","刻俄柏","风笛","傀影","温蒂","早露","玲兰",
"棘刺","森蚺","史尔特尔","瑕光","泥岩","山","空弦","嵯峨","凯尔希","帕拉斯","水月","琴柳","远牙","焰尾"};
if(huan==0){
cout<<l[x];
printf("[★★★★★★]");
huan++;
}
else if(huan==4)
{
cout<<' '<<l[x];
printf("[★★★★★★]\n");
huan=0;
}
else
{
cout<<' '<<l[x];
printf("[★★★★★★]");
huan++;
}
return;
}
void choujiang(int x)
{
double dd;
dd=double(j)*4+double(k);
dd/=n;
printf("欧皇鉴定啦~~~~~~\n经过检验,你是"); //随便搞得-v-
if(dd>=0.4)
{
cout<<"超级欧皇！！ 可恶啊，我好羡慕！！！，你已经超越了人类了（不做人了）\n";
}
else if(dd>=0.3&&dd<0.4)
{
cout<<"欧皇！ 真牛逼，你已经超过了绝大多数人，真贵族血脉！\n";
}
else if(dd>=0.2&&dd<0.3)
{
cout<<"欧洲人！ 你还挺幸运的，慕了。\n";
}
else if(dd>=0.09&&dd<0.2)
{
cout<<"亚洲人， 我们都是黄皮肤。\n";
}
else if(dd>0.07&&dd<0.09)
{
cout<<"非洲人 ，诶呀，没事的，氪就解决了";
}
else
{
cout<<"非酋！！！哈哈哈哈哈哈哈，回家吧酋长！\n";
}
return;
}
int main()
{
printf("这里是简简单单的明日方舟抽奖系统，有保底机制。\n"
"请输入你想抽奖的次数(输入0结束)\n");
while(scanf("%d",&n)!=EOF)
{
//string str="weeee";
// cout<<str.substr();
shichou=0;
ten=0,wushi=0;
j=0,k=0,r=0,huan=0,h=0;
if(n==0)break;
else if(n<0){
printf("请输入正确的数字：\n");
continue;
}
else
{
for(int i=0;i<n;i++)
{
a=rand()%1000;
if(a>=0&&a<(20+o))
{
liu(0);
j++;
if(shichou==0)shichou=1;
wushi=0;
o=0;
}
else if(a>19+o&&a<100+o)
{
wu(0);
k++;
if(shichou==0)shichou=1;
wushi++;
}
else if(a>99+o&&a<600+o)
{
if(ten==9&&shichou==0)
{
k++;
shichou=1;
wu(0);
wushi++;
}
else
{
si(0);
r++;
wushi++;
ten++;
}
}
else
{
if(ten==9&&shichou==0)
{
shichou=1;
wu(0);
k++;
wushi++;
}
else
{
san(0);
h++;
wushi++;
ten++;
}
}
if(wushi>50)o=(wushi-50)*20;
}
}
cout<<endl;
cout<<"总计："<<endl;
printf("六星总数目：%d 五星总数目：%d 四星总数目：%d 三星总数目：%d。\n",j,k,r,h);
n*=10;
printf("消耗合成玉：%d，折合源石： %.2lf\n\n",n*60,double(n)/3);
n/=10;
if(n<10)
{
printf("\n因为次数不足，所以无法做欧皇鉴定\n是否继续？如继续，请输入次数，反之输入0\n\n");
}
else
{
choujiang(n);
printf("\n是否继续？如继续，请输入次数，反之输入0\n\n");
}
}
return 0;
}挺简单的，如果有优化务必评论，我还在学习，我要学习！！