D. Present- (binary bit operation +)

to sum up

See this problem, have been thinking mathematically, using the formula comes to shove, and then firmly dead inside the

Resolve

Thinking every bit to 1, A traversal odd numbers and every two of the bit is 1
B [J]% (J + 1 << 1), the modulo process useless data, seeking to facilitate binary range,
B [J] = [0, 2 ^{I +. 1} -1]
B [J] + B [+ J. 1] = [0, 2 ^{I + 2} -1]
For example: pos = 2, the bit. 1
NUM + B [J ] may be in the range of 1: 0100-0111 or 1100-1111
so: [2 ^I , 2 ^{I + 1} -1] or [2 ^I +2 ^{I + 1} , 2 ^{I 2 +} -2]
Therefore: [2 ^I , 2 ^{I +. 1} -1 -b [J]] or [2 ^I +2 ^{I +. 1} , 2 ^{I + 2} -2-B [J]]

Topic Link

#include<bits/stdc++.h>
//typedef long long ll;
//#define ull       unsigned long long
//#define int       long long
#define F           first
#define S           second
#define endl        "\n"//<<flush
#define eps         1e-6
#define lowbit(x)   (x&(-x))
#define PI          acos(-1.0)
#define inf         0x3f3f3f3f
#define MAXN        0x7fffffff
#define INF         0x3f3f3f3f3f3f3f3f
#define pa          pair<int,int>
#define ferma(a,b)  pow(a,b-2)
#define pb          push_back
#define all(x)      x.begin(),x.end()
#define memset(a,b) memset(a,b,sizeof(a));
#define IOS         ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
using namespace std;
void file()
{
#ifdef ONLINE_JUDGE
#else
    freopen("cin.txt","r",stdin);
    //  freopen("cout.txt","w",stdout);
#endif
}
int get(int num)
{
    return 1<<num;
}
signed main()
{
    IOS;
    //file();
    int n;
    cin>>n;
    vector<int>a(n),b(n);
    for(auto &it:a)
        cin>>it;
    int ans=0;
    for(int i=0;i<26;++i)
    {
        int mod=get(i+1);
        for(int j=0;j<n;++j)
            b[j]=a[j]%mod;
        long long sum=0;
        sort(all(b));
        for(int j=0;j<n;++j)
        {
            int l=lower_bound(all(b),get(i)-b[j])-b.begin();
            int r=upper_bound(all(b),get(i+1)-1-b[j])-b.begin()-1;
            sum+=r-l+1;
            l=lower_bound(all(b),get(i)+get(i+1)-b[j])-b.begin();
            r=upper_bound(all(b),get(i+2)-2-b[j])-b.begin()-1;
            sum+=r-l+1;
            if((b[j]*2)&get(i))
                sum--;
        }
        if(sum/2%2)
            ans+=get(i);
    }
    cout<<ans<<endl;
    return 0;
}