이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
using namespace std;
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define ub upper_bound
#define lb lower_bound
#define sz(x) (int)x.size()
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(),x.rend()
#define NeedForSpeed ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<ll> vll;
typedef vector<int> vii;
typedef vector<pll> vpll;
typedef vector<pii> vpii;
template<class T> bool umin(T& a, const T& b) {return a > b? a = b, true:false;}
template<class T> bool umax(T& a, const T& b) {return a < b? a = b, true:false;}
int msb(int val){return sizeof(int)*8-__builtin_clzll(val)-1;}
const int N = 1e6+5;
const int mod = 1e9+7;
const ll inf = 1e18;
const ld Pi = acos(-1);
#define MULTI 0
int n, m, s, k, ans, a[N];
int cnt[40], pr[40];
void solve(int t_case){
cin>>n;
for(int i=0;i<n;i++) cin>>a[i];
ll res = 0;
for(int i=0;i<n;i++){
for(int j=0;j<31;j++) cnt[j] += (a[i] >> j & 1);
for(int j=0;j<31;j++) if(a[i] >> j & 1 && !(a[i] >> (j+1) & 1)) pr[j]++;
}
//for(int j=0;j<4;j++) cout<<cnt[j]<<" ";
//cout<<"\n";
res |= (cnt[0] * (n - cnt[0]) & 1);
for(int b=1;b<31;b++){
if((cnt[b] * (n - cnt[b]) + pr[b-1] * (pr[b-1] - 1) / 2 + pr[b-1]) % 2) res |= (1 << b);
}
cout<<res<<"\n";
}
signed main(){
NeedForSpeed
if(!MULTI){
solve(1);
} else{
int t;
cin>>t;
for(int t_case = 1; t_case <= t; t_case++) solve(t_case);
}
return 0;
}
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