程序竞赛中利用 ios_base::sync_with_stdio(false) 与 cin.tie(NULL) 加速 IO

Fast I/O for Competitive Programming

In competitive programming, it is important to read input as fast as possible so we save valuable time.

You must have seen various problem statements saying: “Warning: Large I/O data, be careful with certain languages (though most should be OK if the algorithm is well designed)”. The key for such problems is to use Faster I/O techniques.

It is often recommended to use scanf/printf instead of cin/cout for fast input and output. However, you can still use cin/cout and achieve the same speed as scanf/printf by including the following two lines in your main() function:

ios_base::sync_with_stdio(false);
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It toggles on or off the synchronization of all the C++ standard streams with their corresponding standard C streams if it is called before the program performs its first input or output operation. Adding ios_base::sync_with_stdio (false); (which is true by default) before any I/O operation avoids this synchronization. It is a static member of the function of std::ios_base.

 cin.tie(NULL);
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tie() is a method that simply guarantees the flushing of std::cout before std::cin accepts an input. This is useful for interactive console programs which require the console to be updated constantly but also slows down the program for large I/O. The NULL part just returns a NULL pointer.

Moreover, you can include the standard template library (STL) with a single include:

 #include 
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So your template for competitive programming could look like this:

#include 
using namespace std;

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    return 0;
}
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It is recommended to use cout << “\n”; instead of cout << endl;. endl is slower because it forces a flushing stream, which is usually unnecessary (See this for details). (You’d need to flush if you were writing, say, an interactive progress bar, but not when writing a million lines of data.) Write ‘\n instead of endl.

We can test our input and output methods on the problem INTEST – Enormous Input Teston SPOJ. Before further reading, I would suggest you solve the problem first.
Solution in C++ 4.9.2

Normal I/O: The code below uses cin and cout. The solution gets accepted with a runtime of 2.17 seconds.

// A normal IO example code 
#include  
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; 
} 
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Fast I/O However, we can do better and reduce the runtime a lot by adding two lines. The program below gets accepted with a runtime of 0.41 seconds.

// A fast IO program 
#include  
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; 
} 
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Now, talking about competitive contests like ACM ICPC, Google CodeJam, TopCoder Open, here is an exclusive code to read integers in the fastest way.

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; 
} 
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