need some help for this problem

[mrformatt]

Hello,
Write a program that reads a sequence of numbers (at least three), The program must find out three smallest number
for example :
if we entered { 0.5 , 5 , 6 , 150 , 0.25 , 8 , 8 , 1 ,3 }
then the program should outputs { 0,25 , 0,5 , 1 }.

i wrote the following code but it does not work for all number , when i enter :

How many number in your sequence : 7
Please enter first integer :0.5
Please enter next integer :
1
Please enter next integer :8
Please enter next integer 8
Please enter next integer 8
Please enter next integer 4
Please enter next integer 0.25
The Minimum Smallest 0
The Mid Number is 1
The Max Smallest 4

how can i solve this problem ?

#include <iostream>
using namespace std ;

double Min( double , double , double );//function to calculate minimum number
double Mid( double , double , double );//function to calculate second smallest number
double Max( double , double , double );//function to calculate third smallest number



int main() {
    
    double num1 , num2 , num3 ;
    int seq;
    double th3 , sec , frst ;
    
   
    
    cout << "How many number in your sequence : " ; // here we enter how long our sequence is
    cin >> seq;
    cout << "Please enter first integer :"  ;// first we enter three number for have first three smallest numbers
    cin >> num1;
    cout << "Please enter next integer :"   ;
    cin >> num2;
    cout << "Please enter next integer :"   ;
    cin >> num3;
    
    
    th3 = Max( num1 , num2 , num3 );
    sec = Mid( num1 , num2 , num3 );
    frst = Min( num1 , num2 , num3 );
    
  
    // here we wanna enter further number depending on how long our sequnce is
    for ( int i = 4 ; i <= seq ; i++){
    
    
        int New ;
        cout << "Please enter next integer " ;
        cin >> New ;
    
    
        
        if ( New < frst ){
            frst = New ;
        }
        else if ( New > frst && New < sec ){
        
            sec = New ;
        }
        else if ( New > sec && New < th3 ){
        
            th3 = New ;
        }
    }
    
    
    

    cout << "The Minimum Smallest  " <<  frst << endl ;
    cout << "The Mid Number is   "   <<  sec << endl ;
    cout << "The Max Smallest  "     <<  th3 << endl ;
    
    
    
    return 0;
    
}

double Min( double x , double y , double z){

    double Min = x;
    if(y < Min){
        Min = y;}
    
    if (z < Min){
        Min = z;}
    

    return Min;
}

double Max(double x , double y , double z){

    double Max = x;
    
    if( y > Max){
        Max = y ;}
     if( z > Max ){
        Max = z;}
    return Max;
    

}

double Mid(double x , double y , double z){

    double Mid;
    if ( y > x && y < z ){
    
        Mid = y;
    }
    else if ( y < x && y > z  ){
        Mid = y;
    }
    
    
    else if ( x > y && x < z ){
        Mid = x;}
    
    else if ( x < y && x > z ){
        Mid = x ;}
    
    
    else if ( z > y && z < x ){
        Mid = z;
    }
    else if ( z < y && z > x ){
        Mid = z ;
    }
    
    return Mid ;

}

[Esslercuffi]

consider this

num1 = 9999999; // numbers larger than the expected range of input
num2 = 9999999;
num3 = 9999999;
...

for( int i = 0; i < seq; ++i )
{
	cout << "Enter number: ";
	cin >> temp;

	if( temp <= num1 )
	{
		num3 = num2;
		num2 = num1;
		num1 = temp;
	}
	else if( temp <= num2 )
	{
		num3 = num2;
		num2 = temp;
	}
	else if( temp <= num3 )
		num3 = temp;
}

[TheKingOfTyrants]

First, I'd suggest taking all but the first number using a loop. The code is unecessarily complicated by taking three number and then loop through the rest the user requested. Just make sure the user enters a valid number of preferred entries.

Second, you don't need to even use functions for this particular problem. You'd get the first number, seed frst, sec, and th3 equal to it. You'd then check after each number entered to figure out if it fits somewhere in those three variables. You'd also need to shift your variables over. Let's say you get a new lowest value for frst:

th3 = sec;
sec = frst;
frst = new_lowest_value;