今编的一段码,大家提意见  关于数学周易

本文介绍了一个使用C++实现的程序,该程序能够生成特定条件下的六十四卦序列,并将其存储到文本文件中。通过对不同进制和位数的设定,程序实现了对序列的截肢处理和重复检查。

#include <iostream>
#include <string>
#include <vector>
#include <fstream>
using namespace std;
extern int mi(int,int);
extern bool gaishan(vector<int>,int,int);
extern int jh(vector<int>,int,int);
extern int youtong(vector<int>,vector<int>,int);
extern bool chongfu(vector<vector<int> > vvi,vector<int>,int wei,int changdayin);
int main(){
    for(;;){
    int jinzhi=2;std::cout<<"请输入进制:";std::cin>>jinzhi;
    int wei=6;std::cout<<"请输入位数:"; std::cin>>wei;
    int changdayin=mi(jinzhi,wei);
    //第一个”数组”初始化
         
    vector<int> vi;
    for(int i=0;i<wei;i++){vi.push_back(0);}
    //定义一个向量的向量
    vector<vector<int> > vvi;
    vvi.push_back(vi);
    vector<int>::iterator vi_iterator;
    int number_will_check=1;
    int number_checked=0;
    ofstream outfile("六十四卦顺序_2.txt",ios_base::app);
          if(!outfile)cerr<<"没有地方存储!!!";
           else
           outfile<<endl<<"进制:"<<jinzhi<<";"<<"位:"<<wei<<";"<<endl;
    while(0<number_will_check){//没有到最后,就继续
      
          //在此处加一些“截肢”代码
         vi=*vvi.begin();
         int chang=vi.size();
         int _delet=0;
         int i=chang-wei+1;
         for(;i<changdayin;i++){
            int zt=jh(vi,i,wei);
                     if(zt==3){vvi.erase(vvi.begin());//加0和加1都不可以
                               number_will_check--;
                               _delet=1;
                               goto lb1;
                     }
                     if(zt==2)vi.push_back(0);//加0可以,加1不可以                 
                     if(zt==1){vi.push_back(1);}//加1可以,加0不可以               
                     if(zt==0){                 //加0和加1都可以
                              vi.push_back(1);
                              if(jh(vi,i+1,wei)!=3)
                                 vvi.insert(vvi.begin()+1,vi);
                              vi.pop_back();              
                              vi.push_back(0);
                              number_will_check++;                              
                               }                
                      
           
            }//for
                         //if(chongfu(vvi,vi,wei,changdayin)){vvi.erase(vvi.begin()+number_checked);_delet=1;}
                      lb1:if(!_delet){                     
                               vvi.erase(vvi.begin());
                               number_will_check--;                              
                               number_checked++;                 
                               outfile<<number_checked<<":";
                               for(int i=0;i<changdayin;i++)outfile<<vi[i];
                               outfile<<endl;
                               cout<<"等待检验的:"<<number_will_check<<" ";
                               cout<<"符合条件的:"<<number_checked<<endl;
                            
                            }
            }//外层while
        
           
           
            }//最外层for
   

}
int mi(int i,int j){
    int hehe=1;
    for(int k=0;k<j;k++){
            hehe*=i;
            }
    return hehe;

    }

           


//含有相同的“节”就删除
bool gaishan(vector<int> gaishan_vi,int cixu,int wei){
           vector<int>::iterator j2;
           vector<int>::iterator i=gaishan_vi.begin();
     for(int y=0;y<cixu;y++){
           vector<int>::iterator j1=i;
           i++;
           j2=gaishan_vi.begin();
           for(int x=0;x<cixu;x++){j2++;}
           for(int k=0;k<wei;k++){
                   if((*j1)!=(*j2))break;
                   j1++;
                   j2++;            
                   if(k==(wei-1))return 1;
                   } 
                                         
             }
     return 0;
     }
int jh(vector<int> vi,int cixu,int wei){
            //加0后的状态
            vi.push_back(0);
            bool h0=gaishan(vi,cixu,wei);
            vi.pop_back();
            //加1后的状态
            vi.push_back(1);
            bool h1=gaishan(vi,cixu,wei);
            vi.pop_back();
            return(h1*2+h0);
    }    

 

运行部分结果

进制:2;位:6;
1:0000001000011000101000111001001011001101001111010101110110111111
2:0000001000011000101000111001001011001101001111011011101010111111
3:0000001000011000101000111001001011001101001111110101011101101111
4:0000001000011000101000111001001011001101001111110110111010101111
5:0000001000011000101000111001001011001101010111010011110110111111
6:0000001000011000101000111001001011001101010111010011111101101111
7:0000001000011000101000111001001011001101010111011011110100111111
8:0000001000011000101000111001001011001101010111011011111101001111
9:0000001000011000101000111001001011001101010111101101110100111111
10:0000001000011000101000111001001011001101010111111011011101001111
11:0000001000011000101000111001001011001101101001111010101110111111
12:0000001000011000101000111001001011001101101001111011101010111111
13:0000001000011000101000111001001011001101101001111110101011101111
14:0000001000011000101000111001001011001101101001111110111010101111
15:0000001000011000101000111001001011001101101010111010011110111111
16:0000001000011000101000111001001011001101101010111010011111101111
17:0000001000011000101000111001001011001101101010111011110100111111
18:0000001000011000101000111001001011001101101010111011111101001111

 

运行结果有147M

打都打不开,想分别存到几个文件中,但不知道怎么弄

六自由度机械臂ANN工神经网络设计:正向逆向运动学求解、正向动力学控制、拉格朗日-欧拉法推导逆向动力学方程(Matlab代实现)内容概要:本文档围绕六自由度机械臂的ANN工神经网络设计展开,详细介绍了正向与逆向运动学求解、正向动力学控制以及基于拉格朗日-欧拉法推导逆向动力学方程的理论与Matlab代实现过程。文档还涵盖了PINN物理信息神经网络在微分方程求解、主动噪声控制、天线分析、电动汽车调度、储能优化等多个工程与科研领域的应用案例,并提供了丰富的Matlab/Simulink仿真资源和技术支持方向,体现了其在多学科交叉仿真与优化中的综合性价值。; 适合群:具备一定Matlab程基础,从事机器控制、自动化、智能制造、电力系统或相关工程领域研究的科研员、研究生及工程师。; 使用场景及目标:①掌握六自由度机械臂的运动学与动力学建模方法;②学习工神经网络在复杂非线性系统控制中的应用;③借助Matlab实现动力学方程推导与仿真验证;④拓展至路径规划、优化调度、信号处理等相关课的研究与复现。; 阅读建议:建议按目录顺序系统学习,重点关注机械臂建模与神经网络控制部分的代实现,结合提供的网盘资源进行实践操作,并参考文中列举的优化算法与仿真方法拓展自身研究思路。
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值