Change the position of whole digits?

Question

I need to undo an integer like this

input = 12345

output = 54321

i did it, but it gave the wrong conclusion, for example 5432

#include <iostream>
using namespace std;

int main(){
 int num,i=10;   
 cin>>num;   

 do{
    cout<< (num%i)/ (i/10);
    i *=10;
   }while(num/i!=0);

 return 0;
}

Answer 1

Your loop ends too soon. Change

}while(num/i!=0);

to

}while((num*10)/i!=0);

to get another iteration and your code will work.

Answer 2

Here is the solution

    int num = 12345;
    int new_num = 0;
    while(num > 0)
    {
            new_num = new_num*10 + (num % 10);
            num = num/10;
    }
    cout << new_num << endl;

Answer 3

, .

: 12

:

out: 12% 10 = 2/1 = 2
= 100
test: 12/100 = 0 ( )

.

(num% i)!= num

.

Answer 4

, , ( ?) C. num / i , num < 10 i = 10?

Answer 5

while

while (i<10*num)

Answer 6

, () int, . , . , :

output = 0;

while (input !=0)
    output *= 10
    output += input % 10
    input /= 10
}
print output

, :

std::stringstream buffer;

buffer << input;

cout << std::string(buffer.str().rbegin(), buffer.str().rend());

Answer 7

int _tmain(int argc, _TCHAR* argv[])
{
int x = 1234;
int out = 0;
while (x != 0)
{
    int Res = x % (10 );
    x /= 10;
    out *= 10;
    out +=  Res;
}
cout << out;


} 

Answer 8

. ++. , .

Answer 9

.

#include<iostream>
using namespace std;

int main() {
    int n;
    cin >> n;
    cout << (n%10) << (n/10);    
return 0;
}

Answer 10

int a,b,c,d=0;
cout<<"plz enter the number"<<endl;
cin>>a;
b=a;
do
{
    c=a%10;
    d=(d*10)+c; 
    a=a/10;
}
while(a!=0);
cout<<"The reverse of the number"<<d<<endl;
if(b==d)
{
    cout<<"The entered number is palindom"<<endl;
}
else
{
    cout<<"The entered number is not palindom"<<endl;
}

}

Answer 11

template <typename T>
T reverse(T n, size_t nBits = sizeof(T) * 8)
{
    T reverse = 0;
    auto mask = 1;

    for (auto i = 0; i < nBits; ++i)
    {
        if (n & mask)
        {
            reverse |= (1 << (nBits - i - 1));
        }
        mask <<= 1;
    }

    return reverse;
}

(short, byte, int, long...). nBits .

. . 7 8 = 00000111 → 11100000 7 4 = 0111 → 1110

Answer 12

TestDS {

public static void main(String[] args) {

    System.out.println(recursiveReverse(234));
           System.out.println(recursiveReverse(234 ,0));



}



public static int reverse(int number){

    int reversedNumber = 0;
    int temp = 0;


    while(number > 0){

        //use modulus operator to strip off the last digit
        temp = number%10;

        //create the reversed number
        reversedNumber = reversedNumber * 10 + temp;
        number = number/10;

    }



    return reversedNumber;

}



private static int reversenumber =0;
public static int recursiveReverse(int number){

    if(number <= 0){

        return reversenumber;
    }

    reversenumber = reversenumber*10+(number%10);
    number =number/10;

    return recursiveReverse(number);

}

public static int recursiveReverse(int number , int reversenumber){

    if(number <= 0){

        return reversenumber;
    }

    reversenumber = reversenumber*10+(number%10);
    number =number/10;

    return recursiveReverse(number,reversenumber);

}

}

Answer 13

I made it simple, but it is applicable up to five-digit numbers, but I hope this helps

 #include<iostream>
    using namespace std;
void main()
{ 
    int a,b,c,d,e,f,g,h,i,j;
    cin>>a;
    b=a%10;
    c=a/10;
    d=c%10;
    e=a/100;
    f=e%10;
    g=a/1000;
    h=g%10;
    i=a/10000;
    j=i%10;
    cout<<b<<d<<f<<h<<j;
}`