Fast Input and Output in C++

In competitive programming, it's critical to read input as quickly as possible in order to save time. "Warning: Big I / O data, be aware of certain languages (but most should be correct if the operating system is sufficiently developed)," you have probably read in many of the problem statements. Faster I / O systems are key to solving such problems.

For faster and more intuitive input, scanf / printf is often recommended over cin / cout. However, by setting the following two lines in your main (), you can still use cin / cout and get the same functionality as scanf / printf:

ios_base :: sync_with_stdio ( false ) ;

When called before the system performs its first input or output action, it turns on or off the synchronization of all normal C ++ streams and your normal C-stream. This sync can be avoided by adding ios base :: sync with stdio ( false ); (which is automatically true) before every I / O action. Stable member std :: ios base function.

cin.tie ( NULL ) ;

tie () is a function that merely ensures that std :: cout is flushed before std :: cin accepts a command. This is handy for interactive console applications that require the console to be updated on a regular basis, but it slows down the program when dealing with big amounts of data. The NULL portion just returns a reference to NULL.

Furthermore, the standard template library (STL) may be included with a single include:

#include < bits/stdc++.h >

As an example, your competitive programming template may look like this:

#include < bits/stdc++.h >
using namespace std ; 
int main ( )
{
    ios_base :: sync_with_stdio ( false ) ;
    cin.tie ( NULL ) ;
    return 0 ;
}

It is preferable to use cout "n"; rather than cout endl;. endl is more time consuming since it requires a flushing stream, which is frequently unneeded. (If you are upgrading, for example, a collaborative progress bar, you will need to delete it, but not if you are writing a million data lines.) Instead of 'house, write' n. Problem INTEST - Enormous Input Teston SPOJ, we may test our inputs and output methods. I recommend that you tackle the problem first before continuing.

Example:

// A normal IO example code
#include < bits/stdc++.h >
using namespace std ;
int main ( )
{
	int n , k , t ;
	int cnt = 0 ;
	cin >> n >> k ;
	for ( int i=0 ; i < n ; i++ )
	{
		cin >> t ;
		if ( t % k == 0 )
			cnt++ ;
	}
	cout << cnt << "\n" ;
	return 0 ;
}

OUTPUT:

0
...
Process executed in 2.11 seconds
Press any key to continue.

Explanation

cin and cout are used in the code below. With a runtime of 2.11 seconds, the solution is acceptable.

Another Example:

// A fast IO program
#include < bits/stdc++.h >
using namespace std ; 
int main ( )
{
	// added the two lines below
	ios_base :: sync_with_stdio ( false ) ;
	cin.tie ( NULL ) ; 
	int n , k , t ;
	int cnt = 0 ;
	cin >> n >> k ;
	for ( int i = 0 ; i < n ; i++ )
	{
		cin >> t ;
		if ( t % k == 0 )
			cnt++ ;
	}
	cout << cnt << "\n" ;
	return 0 ;
}

OUTPUT:

0
...
Process executed in 0.22 seconds
Press any key to continue.

Explanation

By adding two lines, we can improve the performance and significantly cut the runtime. With a runtime of 0.22 seconds.

Another Example:

#include < bits/stdc++.h >
using namespace std;
void fastscan ( int &number )
{
	//variable to indicate sign of input number
	bool negative = false ;
	register int c ; 
	number = 0 ; 
	// extract current character from buffer
	c = getchar ( ) ;
	if ( c == '-' )
	{
		// number is negative
		negative = true ; 
		// extract the next character from the buffer
		c = getchar ( ) ;
	}
	// Keep on extracting characters if they are integers
	// i.e ASCII Value lies from '0'( 48 ) to '9' ( 57 )
	for ( ; ( c > 47 && c < 58 ) ; c=getchar ( ) )
		number = number *10 + c - 48 ; 
	// if scanned input has a negative sign, negate the
	// value of the input number
	if ( negative )
		number *= -1 ;
}
// Function Call
int main ( )
{
	int number ;
	fastscan ( number ) ;
	cout << number << "\n" ;
	return 0 ;
}

OUTPUT:

0
...
Process executed in 0.12 seconds
Press any key to continue.

Explanation:

Now, when it comes to competitive challenges such as the ACM ICPC, Google CodeJam, and TopCoder Open, above is a unique code for reading numbers quickly.