Problem:
Score : 100 points

Problem Statement

Given is an integer x that is greater than or equal to 0, and less than or equal to 1.
Output 1 if x is equal to 0, or 0 if x is equal to 1.

Constraints

0 \leq x \leq 1

x is an integer

Input

Input is given from Standard Input in the following format:

x

Output

Print 1 if x is equal to 0, or 0 if x is equal to 1.

Sample Input 1

1

Sample Output 1

0

Sample Input 2

0

Sample Output 2

1

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;
typedef  long long ll;

#define pf printf
#define pi acos(-1.0)
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define pb push_back
#define mp make_pair
#define sz(v) ((int)(v).size())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)
#define PER(i,n) for(int i=(n-1);i>=(0);--i)
#define REPSZ(i,v) REP(i,SZ(v))
#define pll pair <ll, ll>

const ll mxn = 1e5+5;
ll mod = 1e9+7;
ll arr[mxn];
ll brr[mxn];
vector<ll>v;
bitset<mxn>used;
double eps = 1e-7;
ll n,m,k,cont,tc;
void print(vector <ll> &v){cout << v.size() << endl;for(int i = 0; i < v.size(); i++){pf("%lld ", v[i]);}pf("\n");}
void print(vector <pll> &v){ cout << v.size() << endl; for(int i = 0; i < v.size(); i++){pf("%lld %lld\n", v[i].first, v[i].second);}}
void print(set<ll> s){set <ll> :: iterator itr;for (itr = s.begin(); itr != s.end(); ++itr){cout << *itr << " ";}cout << endl;}
void print(double d){cout << fixed << setprecision(2) << d << endl;}
void print (ll *arr){REP(i,n)cout << arr[i] << " "; cout << endl;}

void from_file(void){ freopen("input.txt","r",stdin);freopen("output.txt","w",stdout);}

/* ------------------main section-------------! */
ll Power(ll a, ll b) {
    a %= mod;
    ll res = 1;
    while (b > 0) {
        if (b & 1)
            res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return res;
}

ll NcR(int n, int r) 
{ 
  
    ll p = 1, k = 1; 
    if (n - r < r) 
        r = n - r; 
  
    if (r != 0) { 
        while (r) { 
            p *= n; 
            k *= r; 
  
            ll m = __gcd(p, k); 
            p /= m; 
            k /= m; 
  
            n--; 
            r--; 
        } 
    } 
  
    else
        p = 1; 
    return p;
} 

void solve()
{
    
}

void run() {
    fastio;
    bool x;
    cin >> x;
    cout << (x^1) << endl;
}

int main() {
    #ifndef ONLINE_JUDGE
       from_file();
    #endif
    run();
    return 0;
}