HackerRank Basic Data Types solution in c++ programming

In this HackerRank Basic Data Types problem solution in the c++ programming language, Some C++ data types, their format specifiers, and their most common bit widths are as follows:

Int (“%d”): 32 Bit integer

Long (“%ld”): 64 bit integer

Char (“%c”): Character type

Float (“%f”): 32 bit real value

Double (“%lf”): 64 bit real value

Reading

To read a data type, use the following syntax:

scanf(“`format_specifier`”, &val)

For example, to read a character followed by a double:

char ch;

double d;

scanf(“%c %lf”, &ch, &d);

For the moment, we can ignore the spacing between format specifiers.

Printing

To print a data type, use the following syntax:

printf(“`format_specifier`”, val)

For example, to print a character followed by a double:

char ch = ‘d’;

double d = 234.432;

printf(“%c %lf”, ch, d);

HackerRank Basic Data Types problem solution in c++ programming.

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int a; long b; char c; float d; double e;
    cin>>a>>b>>c>>d>>e;
    cout<<a<<"n"<<b<<"n"<<c<<"n";
    cout<<fixed<<setprecision(3)<<d<<"n";
    cout<<fixed<<setprecision(9)<<e<<"n";
    return 0;
}

Second solution

#include <iostream>
#include <cstdio>
#include <iomanip> 
using namespace std;

int main() {
    
    int a;
    long b;
    char c;
    float d;
    double e;
    cin >> a >> b >> c >> d >> e;
    cout<< a << 'n' << b << 'n' << c << 'n';
    cout << std::fixed << std::setprecision(3) << d << 'n';
    cout << std::fixed << std::setprecision(9) << e << 'n';
    
    return 0;
}

Third solution

#include <iostream>
#include <cstdio>
using namespace std;

int main() {
    // Complete the code.
    int i;
    long l;
    char c;
    float f;
    double d;
    scanf("%d %ld %c %f %lf" , &i, &l, &c, &f, &d);
    printf("%dn%ldn%cn%.3fn%.9lfn", i,l,c,f,d);

    return 0;
}