For beginners

• If you are new to learning programming and do not know where to start, please try Problem A "Welcome to AtCoder" from practice contest. There you can find a sample code for each language.
  
• Also, if you are not familiar with problems in programming contests, we recommend you to try some problems in "AtCoder Beginners Selection".
  
• 「C++入門 AtCoder Programming Guide for beginners (APG4b)」 is a C++ tutorial for competitive programmers. Sadly, this is only in Japanese too.
  
• 「Python入門 AtCoder Programming Guide for beginners (APG4bPython)」 is a Python tutorial for competitive programmers. Again, this is only in Japanese.
  

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Original proposer: admin

The problem can be answered by computing the \(N\)-th power of \(2\) and \(2 \times N\), and finding the difference between them.
The multiplication and subtraction seems relatively easy, but how can we compute the \(N\)-th power of \(2\)?

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Solution \(1\): use a for loop

The \(N\)-th power of \(2\) can be computed using a for loop.

Sample code (C++):

#include<bits/stdc++.h>

using namespace std;

int main(){
  int n;
  cin >> n;
  int p=1;
  for(int i=0;i<n;i++){
    p=p*2;
  }
  cout << p-2*n << "\n";
  return 0;
}

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Solution \(2\): use the fast exponentiation algorithm

One may also adopt the fast exponentiation, which is an algorithm to find powers fast.

Sample code (C++):

#include<bits/stdc++.h>

using namespace std;

int main(){
  int n;
  cin >> n;
  int a=1,b=2,p=n;
  while(p>0){
    if(p%2==1){ a*=b; }
    b*=b;
    p/=2;
  }
  cout << a-2*n << "\n";
  return 0;
}

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Solution \(3\): use a standard function

For example, in Python one can compute powers using **.

Sample code (Python):

n = int(input())
print(2**n-2*n)

In C++ also, this problem can be solved with pow function, but note that the result of pow(int,int) against int-type values results in a double-type value, which requires an additional caution as it may lead to precision errors or output format violation.

Sample code (C++):

#include<bits/stdc++.h>

using namespace std;

int main(){
  int n;
  cin >> n;
  cout << pow(2,n)-2*n << "\n";
  return 0;
}

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Solution \(4\): use the bitwise-shift operator

With bitwise-shift operator a<<k, one can obtain the value \(a \times 2^k\) (as long as an overflow does not occur).

Sample code (C++):

#include<bits/stdc++.h>

using namespace std;

int main(){
  int n;
  cin >> n;
  cout << (1<<n)-2*n << "\n";
  return 0;
}