Skip to main content

Faster I/O For CP

 

Faster I/O


You may have come across programming challenges where you are asked to use “faster I/O”. For those of you who have been in the sport programming arena for a while, you already know why you need to do this. This instruction in a challenge is the telltale sign that the test cases contains large input/output data.

How Do You Use Faster I/O?

The answer to this depends on which programming language you are using.

C++

If you are using C++, you can easily switch to faster I/O by adding the following two lines at the beginning of your main() function:

ios_base::sync_with_stdio(false);
cin.tie(NULL);

You must call ios_base::sync_with_stdio(false) before performing any input/output operation. And so, it is a good idea to do it at the beginning of your main function. The effect doing it later is implementation-specific and may not work as expected.

When this function is called with false, it disables synchronization of C++ streams with C streams after each input/output operation.

In addition to this, you can use '\n' at the end of a cout instead of endl:

cout << x << '\n';

Alternatively, you can use printf and scanf in C++ by including “stdio.h”.

printf("%d\n", x);

Further Reading

Comments

Post a Comment

Popular posts from this blog

NT Part 6: Sieve of Eratosthenes(Details)

  About 2300 years ago from today, famous Greek mathematician Euclid proved that there are an infinite number of prime numbers. Since then people have been searching for these prime numbers. In 1849, one of the greatest mathematician of all time, Carl Fredrick Gauss, had identified almost all of the prime numbers within the first 3 hundred thousand whole numbers. In the age of computers, we can find large prime numbers in the blink of an eye. But to do that, we need to know a bit of programming and a 2000 year old algorithm. By the end of this tutorial, you will be able to figure out a solution on your own to Gauss’s problem. What is a Prime Number? A prime number is an integer number that is divisible by 1 and the number itself only. For example, 7 is divisible by 1 and 7 only. But 6 is not a prime number because 6 is be divisible by 2 and 3 as well. It is worth mentioning that 1 itself is not a prime number. Now if you are asked to determine if a number is a prime number, you can...
Post a Comment
Read more

Binary Search & The Kahini!

  Suppose you are given an array of numbers and you need to find whether the number X is in the array. The straight forward approach to do that would be to run a loop and compare each number in the array with X. This is what the pseudo code would look like: function findNumber ( A , x ) for i = 0 to length of array A if A [ i ] == x return true return false For example, let’s assume you are given this array: [7, 14, 17, 21, 35, 60, 92, 121, 155] And, you were looking for the number 92. The loop would start checking each of the number in the array against 92 and would find after iterating through the array 7 times. The worst case time complexity of this approach would be O(n). However, whenever the array itself is in sorted order (like in this case where the numbers are increasing in value from left to right of the array), you can perform something that is much more efficient: binary search. The prerequisite of binary search is that the array should be sor...
Post a Comment
Read more

NT Part 3: Playing with Modulars, and Euler Phi Function

  Quick stuff first, fast exponentiation in logarithm time. Let us calculate   in modular   in  . It uses binary expansion of  ,  and is very very straightforward. ll exp ( ll x, ll n ) { if ( n == 0 ) return 1 ; if ( n == 1 ) return x ; if ( n % 2 == 0 ) return exp ( ( x * x ) % mod,n / 2 ) ; if ( n % 2 == 1 ) return ( x * exp ( ( x * x ) % mod,n / 2 ) ) % mod ; } Now, let us talk about modular inverses. By using Extended Euclidean Algorithm, we can get the inverse of   modulo  . #include <iostream> int inv ( int a, int m ) { int temp = m, q, t, u = 0 , v = 1 ; if ( m == 1 ) return 0 ; while ( a > 1 ) { q = a / m ; t = m ; m = a % m ; a = t ; t = u ; u = v - q * u ; v = t ; } if ( v < 0 ) v + = temp ; return v ; } int main ( void ) { int a, m ; std :: cin ...
Post a Comment
Read more