复制粘贴，看好戏，密码114514（有点浅臭）

#include<bits/stdc++.h>
#define   PI     2*acos(0.0)
#include <stdio.h>
#include <conio.h>
#include <string.h>

using namespace std;
int main(){
    int zz=1;
	int  password,password1,password2;
    password =123;
    int nn = 5,uu;
	cout<<"请输入密码（提示：仅有两次输入机会）："<<endl;
    cin>>password1;
    while(nn){
	    if(password1==password){
	    	cout<<"密码正确！"<<endl<<endl<<endl<<endl; 
	    	cout<<"提示：如果输入1，采用中文执行程序，如果输入2，全程采用英文执行程序。注意注意！！！！！：一旦选定语言，一直都不可更改"<<endl; 
	    	cout<<"转换语言后可能会有错位的情况，可从头看起介绍"<<endl<<endl<<endl; 
    		cin>>uu;
    		if(uu==2){
		    cout<<"I'm a Minimalist Sequential Computing Robot (Chinese and English Version)!"<<endl;
		     
		    cout<<"number 21"<<endl;
		    
			cout<<"The program you are working on now is a beta version of me, and you are welcome to take the test on me!"<<endl;
		 
			cout<<"As a robot, I care a lot about order"<<endl;
		
			cout<<"The order in which I calculate the parts is to add, subtract, multiply, divide, and find the remainder."<<endl;
		
			cout<<"Enter a number, press the spacebar or enter the second number, enter the third number in the same steps, and finally press enter, and I can perform the corresponding operation!"<<endl;
		
			cout<<"As the saying goes, no one is perfect, and the machine is not complete, here are my two shortcomings!"<<endl;
		
			cout<<"Disadvantage 1: The order of operations is calculated strictly according to addition, subtraction, multiplication and division, and cannot skip steps (but you can enter three numbers perfunctory this operation)!"<<endl;
		
			cout<<"Disadvantage 2: I can only perform operations between three numbers at most, if I only perform operations on two numbers, the last number is added and subtracted to write 0, and multiplication and division are written to write 1. I can't mix addition and subtraction and there is no parenthesis mechanism!"<<endl;
		
			long long int a,b,c,d,e,f,i,j;
	   	    long double g,h;
	     
			cout<<"Start adding now!"<<endl; 
	  	    long long int hh;
	 	    cin>>a>>b>>hh;
		    cout<<a+b+hh<<endl<<endl;
	    
			cout<<"Start subtracting now!"<<endl;
			long long int ii; 
			cin>>c>>d>>ii;
			cout<<c-d-ii<<endl<<endl;
		 
			cout<<"Start performing multiplication now! Note that once one of the three multipliers is 0, the overall result is 0"<<endl;
			long long int jj; 
			cin>>e>>f>>jj;
			cout<<e*f*jj<<endl<<endl;
		
			cout<<"Now it's division!"<<endl;
			long long int kk;
		
			cin>>g>>h>>kk;
		
			if(kk||g||h==0){  
				cout<<"This is not allowed, and the following jumps to the remainder section"<<endl<<endl;
			}
			else{
				if(kk=1){
				    cout<<g<<endl<<endl;
			    } 
			    else{
			    	cout<<g/h/kk<<endl<<endl;
				}
			}
		
		
		
			cout<<"Now it's an extra operation, find the remainder!"<<endl;
			cin>>i>>j;
			cout<<i%j<<endl<<endl;
		
			cout<<"This calculation ends"<<endl<<endl;
		
			cout<<"Now please enter a number to represent the number of seconds, and I will convert it into a digital watch form, instantly: minutes: seconds"<<endl;
			long long int t,n,m,s;
			cin>>t;
			n=t/3600;
			m=(t-3600*n)/60;
			s=t-3600*n-60*m;
			if(n<10)cout<<0;
		    cout<<n<<":";
		    if(m<10)cout<<0;
		    cout<<m<<":";
		    if(s<10)cout<<0;
		    cout<<s<<endl<<endl;
	    
		    cout<<"Judging whether the three sides can form a triangle is a necessary learning step in programming, and the judgment procedure is now executed!"<<endl;
			long long int k,l,u;
			cin>>k>>l>>u;
			if(k+l>u&&k+u>l&&l+u>k){
				cout<<"Yes"<<endl<<endl;
			}
			else{
				cout<<"No"<<endl<<endl;
			}
			
			
			cout<<"All aspects of the data of the circle are expansive common problems, enter the radius below, I will give the answer to the circumference and area of the circle!"<<endl;
			long long int r;
			cin>>r;
			cout<<PI*r*r*PI<<endl;
			cout<<2*PI*r<<endl<<endl; 
			
			cout<<"Calculating the factorial of n has several 0s at the end of the very common Arcane problem, and the factorial answer to n is given below!"<<endl;
		    unsigned int aa;
		    unsigned long long factorial = 1;
		    cin >> aa;
		    for(int cc=1;cc<=aa;++cc){
		        factorial *= cc;
		    }
		    cout <<factorial<<endl<<endl;  
			
		 
			cout<<"Calculate the equal difference string and we use the formula, you can now use this program to prove your answer"<<endl;
			
			cout<<"Input format: Enter two numbers, the first number represents the first term of the string, the second number represents the last term of the string, and the third number represents the tolerance of the string"<<endl;
			
			cout<<"For example, enter n and a and m, that is, calculate the sum of a string of numbers with the first term n, the last term a, and the tolerance m"<<endl;
			
		    long long int bb,sum=0,dd,ll,mm;//ll=首项，dd=末项，mm=公差 
			cin>>ll>>dd>>mm;
			for(bb=ll;bb<=dd;bb=bb+mm){
				sum+=bb;
			}
			cout<<sum<<endl<<endl;
			
			
		    cout<<"The problem of counting averages is very common in life, and the following program can solve most of them! Input format: first enter a number n to calculate the number of units to calculate the average, and then enter the same number as n, I can calculate their average, such as: first enter 5, then enter 5 numbers, I will output their average"<<endl;
		    long long int v,w;
			double y=0,x;
			cin>>w;
			for(v=1;v<=w;v++){
				cin>>x;
				y=y+x;
			}
			y=y/w;
			cout<<setprecision(10)<<fixed<<y<<endl;
			
			
			cout<<"Presumably, you must be worried about whether certain numbers are prime numbers. Now, please enter a number and let me tell you if it is prime or not!"<<endl;
			long long int ee,ff,gg=0;
		    cin>>ee;
		    if(ee==1){
		        cout<<"No";
		        return 0;
	        }
	        else{
	            for(ff=2;ff<=sqrt(ee);ff++){
	           	 if(ee%ff==0){
	               	 gg=1;
	               	 break;
	            	}
	       	    }
	            if(gg==0){
	        	    cout<<"Yes";
		    	} 
	            else {
	            	cout<<"No";
			    }
	            }
		            return 0;
	        }
	        else{
		        cout<<"我是极简顺序计算机器人(中英两语版)！"<<endl;
			    
				cout<<"编号021" <<endl;
			
				cout<<"我计算部分的顺序是加，减，乘，除，求余。"<<endl;
			
				cout<<"输入一个数后按空格键或enter键输入第二个数后，以同样的步骤输入第三个数，最后按enter，我就能执行相应运算啦！"<<endl;
			
				cout<<"俗话说人无完人，机器也无完机器，下面是我两个缺点！"<<endl;
			
				cout<<"缺点1：运算顺序是严格按照加减乘除计算的，不能跳步（但是可以随便输入三个数敷衍这一运算）！"<<endl;
			
				cout<<"缺点2：我最多只能进行三个数之间的运算，若只进行两个数的运算，最后一个数加减法写0，乘除法写1。我不能进行加减混合运算且没有括号机制！"<<endl;
			
				long long int a,b,c,d,e,f,i,j;
		   	    long double g,h;
		    
				cout<<"现在开始执行加法运算！"<<endl; 
		  	    long long int hh;
		 	    cin>>a>>b>>hh;
			    cout<<a+b+hh<<endl<<endl;
		    
				cout<<"现在开始执行减法运算！" <<endl;
				long long int ii; 
				cin>>c>>d>>ii;
				cout<<c-d-ii<<endl<<endl;
			
				cout<<"现在开始执行乘法运算！注意，三个乘数中一旦有一个是0，整个结果就是0"<<endl;
				long long int jj; 
				cin>>e>>f>>jj;
				cout<<e*f*jj<<endl<<endl;
			
				cout<<"现在进行的是除法运算！"<<endl;
				long long int kk;
			
				cin>>g>>h>>kk;
			
				cout<<g/h/kk;
			
			
			
				cout<<"现在进行的是额外操作，求余数！"<<endl;
				cin>>i>>j;
				cout<<i%j<<endl<<endl;
			
				cout<<"本次计算结束"<<endl<<endl;

			
				cout<<"现在请您输入一个数表示秒数，我会把他转换成电子表形式，即时：分：秒"<<endl<<endl; 
				long long int t,n,m,s;
				cin>>t;
				n=t/3600;
				m=(t-3600*n)/60;
				s=t-3600*n-60*m;
				if(n<10)cout<<0;
			    cout<<n<<":";
			    if(m<10)cout<<0;
			    cout<<m<<":";
			    if(s<10)cout<<0;
			    cout<<s<<endl<<endl;
		    
			    cout<<"判断三条边能否组成三角形是编程必要学习一步,现执行判断程序！"<<endl<<endl;
				long long int k,l,u;
				cin>>k>>l>>u;
				if(k+l>u&&k+u>l&&l+u>k){
					cout<<"Yes"<<endl<<endl;
				}
				else{
					cout<<"No"<<endl<<endl;
				}
				
				
				
				cout<<"算圆的各方面数据是拓展型常见问题，下面输入半径，我会给出圆周长和面积的答案！"<<endl<<endl;
				long long int r;
				cin>>r;
				cout<<PI*r*r*PI<<endl;
				cout<<2*PI*r<<endl<<endl; 
				
				cout<<"计算n的阶乘末尾有几个0是很常见的奥数问题，下面给出n的阶乘答案!"<<endl<<endl;
			    unsigned int aa;
			    unsigned long long factorial = 1;
			    cin >> aa;
			    for(int cc=1;cc<=aa;++cc){
			        factorial *= cc;
			    }
			    cout <<factorial<<endl<<endl;  
				
			 
				cout<<"计算等差数串的和我们用公式，现可以用此程序证明你的答案"<<endl;
				
				cout<<"输入格式：输入两个数，第一个数表示数串的首项，第二个数表示数串的末项，第三个数表示数串的公差"<<endl;
				
				cout<<"如：输入n和a和m，即表示计算一个首项为n,末项为a，公差为m的数串的和"<<endl<<endl; 
				
			    long long int bb,sum=0,dd,ll,mm;//ll=首项，dd=末项，mm=公差 
				cin>>ll>>dd>>mm;
				for(bb=ll;bb<=dd;bb=bb+mm){
					sum+=bb;
				}
				cout<<sum<<endl<<endl;
				
				
			    cout<<"算平均数的问题在生活中非常常见，下面的程序可以解决大部分问题！输入格式：先输入一个数n表示要计算平均数的单位数，再输入和n一样数量的数，我就可计算它们的平均数，如：先输入5，再输入5个数，我便会输出他们的平均数"<<endl<<endl;
			    long long int v,w;
				double y=0,x;
				cin>>w;
				for(v=1;v<=w;v++){
					cin>>x;
					y=y+x;
				}
				y=y/w;
				cout<<setprecision(10)<<fixed<<y<<endl;
				
				
				cout<<"想必你一定会为某些数是否是质数这个问题而发愁吧。现在，请输入一个数，让我来给您判断它是不是质数！"<<endl<<endl;
				long long int ee,ff,gg=0;
			    cin>>ee;
			    if(ee==1){
			        cout<<"No";
			        return 0;
		        }
			}
			}
		else{
	        cout<<"你输入的密码不正确，请重新输入："<<endl;
	        cin>>password2;
	        if(password2==password){
        	cout<<"密码正确！"<<endl<<endl<<endl<<endl; 
	    	cout<<"提示：如果输入1，采用中文执行程序，如果输入2，全程采用英文执行程序。注意注意！！！！！：一旦选定语言，一直都不可更改"<<endl; 
	    	cout<<"转换语言后可能会有错位的情况，可从头看起介绍"<<endl<<endl<<endl; 
    		cin>>uu;
	    	if(uu==2){
			    cout<<"I'm a Minimalist Sequential Computing Robot (Chinese and English Version)!"<<endl;
			     
			    cout<<"number 21"<<endl;
			    
				cout<<"The program you are working on now is a beta version of me, and you are welcome to take the test on me!"<<endl;
			 
				cout<<"As a robot, I care a lot about order"<<endl;
			
				cout<<"The order in which I calculate the parts is to add, subtract, multiply, divide, and find the remainder."<<endl;
			
				cout<<"Enter a number, press the spacebar or enter the second number, enter the third number in the same steps, and finally press enter, and I can perform the corresponding operation!"<<endl;
			
				cout<<"As the saying goes, no one is perfect, and the machine is not complete, here are my two shortcomings!"<<endl;
			
				cout<<"Disadvantage 1: The order of operations is calculated strictly according to addition, subtraction, multiplication and division, and cannot skip steps (but you can enter three numbers perfunctory this operation)!"<<endl;
			
				cout<<"Disadvantage 2: I can only perform operations between three numbers at most, if I only perform operations on two numbers, the last number is added and subtracted to write 0, and multiplication and division are written to write 1. I can't mix addition and subtraction and there is no parenthesis mechanism!"<<endl;
			
				long long int a,b,c,d,e,f,i,j;
		   	    long double g,h;
		     
				cout<<"Start adding now!"<<endl; 
		  	    long long int hh;
		 	    cin>>a>>b>>hh;
			    cout<<a+b+hh<<endl<<endl;
		    
				cout<<"Start subtracting now!"<<endl;
				long long int ii; 
				cin>>c>>d>>ii;
				cout<<c-d-ii<<endl<<endl;
			 
				cout<<"Start performing multiplication now! Note that once one of the three multipliers is 0, the overall result is 0"<<endl;
				long long int jj; 
				cin>>e>>f>>jj;
				cout<<e*f*jj<<endl<<endl;
			
				cout<<"Now it's division!"<<endl;
				long long int kk;
			
				cin>>g>>h>>kk;
                cout<<g/h/kk;
                
				cout<<"Now it's an extra operation, find the remainder!"<<endl;
				cin>>i>>j;
				cout<<i%j<<endl<<endl;
			
				cout<<"This calculation ends"<<endl<<endl;
			
				cout<<"Now please enter a number to represent the number of seconds, and I will convert it into a digital watch form, instantly: minutes: seconds"<<endl;
				long long int t,n,m,s;
				cin>>t;
				n=t/3600;
				m=(t-3600*n)/60;
				s=t-3600*n-60*m;
				if(n<10)cout<<0;
			    cout<<n<<":";
			    if(m<10)cout<<0;
			    cout<<m<<":";
			    if(s<10)cout<<0;
			    cout<<s<<endl<<endl;
		    
			    cout<<"Judging whether the three sides can form a triangle is a necessary learning step in programming, and the judgment procedure is now executed!"<<endl;
				long long int k,l,u;
				cin>>k>>l>>u;
				if(k+l>u&&k+u>l&&l+u>k){
					cout<<"Yes"<<endl<<endl;
				}
				else{
					cout<<"No"<<endl<<endl;
				}
				
				
				cout<<"All aspects of the data of the circle are expansive common problems, enter the radius below, I will give the answer to the circumference and area of the circle!"<<endl;
				long long int r;
				cin>>r;
				cout<<PI*r*r*PI<<endl;
				cout<<2*PI*r<<endl<<endl; 
				
				cout<<"Calculating the factorial of n has several 0s at the end of the very common Arcane problem, and the factorial answer to n is given below!"<<endl;
			    unsigned int aa;
			    unsigned long long factorial = 1;
			    cin >> aa;
			    for(int cc=1;cc<=aa;++cc){
			        factorial *= cc;
			    }
			    cout <<factorial<<endl<<endl;  
				
			 
				cout<<"Calculate the equal difference string and we use the formula, you can now use this program to prove your answer"<<endl;
				
				cout<<"Input format: Enter two numbers, the first number represents the first term of the string, the second number represents the last term of the string, and the third number represents the tolerance of the string"<<endl;
				
				cout<<"For example, enter n and a and m, that is, calculate the sum of a string of numbers with the first term n, the last term a, and the tolerance m"<<endl;
				
			    long long int bb,sum=0,dd,ll,mm;//ll=首项，dd=末项，mm=公差 
				cin>>ll>>dd>>mm;
				for(bb=ll;bb<=dd;bb=bb+mm){
					sum+=bb;
				}
				cout<<sum<<endl<<endl;
				
				
			    cout<<"The problem of counting averages is very common in life, and the following program can solve most of them! Input format: first enter a number n to calculate the number of units to calculate the average, and then enter the same number as n, I can calculate their average, such as: first enter 5, then enter 5 numbers, I will output their average"<<endl;
			    long long int v,w;
				double y=0,x;
				cin>>w;
				for(v=1;v<=w;v++){
					cin>>x;
					y=y+x;
				}
				y=y/w;
				cout<<setprecision(10)<<fixed<<y<<endl;
				
				
				cout<<"Presumably, you must be worried about whether certain numbers are prime numbers. Now, please enter a number and let me tell you if it is prime or not!"<<endl;
				long long int ee,ff,gg=0;
			    cin>>ee;
			    if(ee==1){
			        cout<<"No";
			        return 0;
		        }
		        else{
			            for(ff=2;ff<=sqrt(ee);ff++){
			           	 if(ee%ff==0){
			               	gg=1;
			               	break;
			            	}
			       	    }
			            if(gg==0){
			        	    cout<<"Yes";
				    	} 
			            else {
			            	cout<<"No";
					    }
		            }
			            return 0;
		        }
	        else{
		        cout<<"我是极简顺序计算机器人(中英两语版)！"<<endl;
			    
				cout<<"编号021" <<endl;
			
				cout<<"我计算部分的顺序是加，减，乘，除，求余。"<<endl;
			
				cout<<"输入一个数后按空格键或enter键输入第二个数后，以同样的步骤输入第三个数，最后按enter，我就能执行相应运算啦！"<<endl;
			
				cout<<"俗话说人无完人，机器也无完机器，下面是我两个缺点！"<<endl;
			
				cout<<"缺点1：运算顺序是严格按照加减乘除计算的，不能跳步（但是可以随便输入三个数敷衍这一运算）！"<<endl;
			
				cout<<"缺点2：我最多只能进行三个数之间的运算，若只进行两个数的运算，最后一个数加减法写0，乘除法写1。我不能进行加减混合运算且没有括号机制！"<<endl;
			
				long long int a,b,c,d,e,f,i,j;
		   	    long double g,h;
		    
				cout<<"现在开始执行加法运算！"<<endl; 
		  	    long long int hh;
		 	    cin>>a>>b>>hh;
			    cout<<a+b+hh<<endl<<endl;
		    
				cout<<"现在开始执行减法运算！" <<endl;
				long long int ii; 
				cin>>c>>d>>ii;
				cout<<c-d-ii<<endl<<endl;
			
				cout<<"现在开始执行乘法运算！注意，三个乘数中一旦有一个是0，整个结果就是0"<<endl;
				long long int jj; 
				cin>>e>>f>>jj;
				cout<<e*f*jj<<endl<<endl;
			
				cout<<"现在进行的是除法运算！"<<endl;
				long long int kk;
			
				cin>>g>>h>>kk;
			
				cout<<g/h/kk;
			
			
			
				cout<<"现在进行的是额外操作，求余数！"<<endl;
				cin>>i>>j;
				cout<<i%j<<endl<<endl;
			
				cout<<"本次计算结束"<<endl<<endl;

			
				cout<<"现在请您输入一个数表示秒数，我会把他转换成电子表形式，即时：分：秒"<<endl<<endl; 
				long long int t,n,m,s;
				cin>>t;
				n=t/3600;
				m=(t-3600*n)/60;
				s=t-3600*n-60*m;
				if(n<10)cout<<0;
			    cout<<n<<":";
			    if(m<10)cout<<0;
			    cout<<m<<":";
			    if(s<10)cout<<0;
			    cout<<s<<endl<<endl;
		    
			    cout<<"判断三条边能否组成三角形是编程必要学习一步,现执行判断程序！"<<endl<<endl;
				long long int k,l,u;
				cin>>k>>l>>u;
				if(k+l>u&&k+u>l&&l+u>k){
					cout<<"Yes"<<endl<<endl;
				}
				else{
					cout<<"No"<<endl<<endl;
				}
				
				
				
				cout<<"算圆的各方面数据是拓展型常见问题，下面输入半径，我会给出圆周长和面积的答案！"<<endl<<endl;
				long long int r;
				cin>>r;
				cout<<PI*r*r*PI<<endl;
				cout<<2*PI*r<<endl<<endl; 
				
				cout<<"计算n的阶乘末尾有几个0是很常见的奥数问题，下面给出n的阶乘答案!"<<endl<<endl;
			    unsigned int aa;
			    unsigned long long factorial = 1;
			    cin >> aa;
			    for(int cc=1;cc<=aa;++cc){
			        factorial *= cc;
			    }
			    cout <<factorial<<endl<<endl;  
				
			 
				cout<<"计算等差数串的和我们用公式，现可以用此程序证明你的答案"<<endl;
				
				cout<<"输入格式：输入两个数，第一个数表示数串的首项，第二个数表示数串的末项，第三个数表示数串的公差"<<endl;
				
				cout<<"如：输入n和a和m，即表示计算一个首项为n,末项为a，公差为m的数串的和"<<endl<<endl; 
				
			    long long int bb,sum=0,dd,ll,mm;//ll=首项，dd=末项，mm=公差 
				cin>>ll>>dd>>mm;
				for(bb=ll;bb<=dd;bb=bb+mm){
					sum+=bb;
				}
				cout<<sum<<endl<<endl;
				
				
			    cout<<"算平均数的问题在生活中非常常见，下面的程序可以解决大部分问题！输入格式：先输入一个数n表示要计算平均数的单位数，再输入和n一样数量的数，我就可计算它们的平均数，如：先输入5，再输入5个数，我便会输出他们的平均数"<<endl<<endl;
			    long long int v,w;
				double y=0,x;
				cin>>w;
				for(v=1;v<=w;v++){
					cin>>x;
					y=y+x;
				}
				y=y/w;
				cout<<setprecision(10)<<fixed<<y<<endl;
				
				
				cout<<"想必你一定会为某些数是否是质数这个问题而发愁吧。现在，请输入一个数，让我来给您判断它是不是质数！"<<endl<<endl;
				long long int ee,ff,gg=0;
			    cin>>ee;
			    if(ee==1){
			        cout<<"No";
			        return 0;
		        }
			}
	    }
		else{
			cout<<"你已经连续输错密码两次，系统将退出！";
			return 0;	
		}	
		}
	}
}