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Codeforces Round #626 (Div. 2, based on Moscow Open Olympiad in Informatics) D.Present

Catherine received an array of integers as a gift for March 8. Eventually she grew bored with it, and she started calculated various useless characteristics for it. She succeeded to do it for each one she came up with. But when she came up with another one — xor of all pairwise sums of elements in the array, she realized that she couldn’t compute it for a very large array, thus she asked for your help. Can you do it? Formally, you need to compute

(a1+a2)⊕(a1+a3)⊕…⊕(a1+an)⊕(a2+a3)⊕…⊕(a2+an)…⊕(an−1+an)
Here x⊕y is a bitwise XOR operation (i.e. x ^ y in many modern programming languages).

Input

The first line contains a single integer n (2≤n≤400000) — the number of integers in the array.

The second line contains integers a1,a2,…,an (1≤ai≤107).

Output

Print a single integer — xor of all pairwise sums of integers in the given array.

Examples

input

2
1 2

output

3

input

3
1 2 3

output

2

借鉴了大神的想法，对答案的每一位二分即可，AC代码如下：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=4e5+5;
int a[N],b[N],n,m,k,ans=0;
int main(){
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>a[i];
    }
    for(int i=0;i<26;i++){
        int mod=1<<(i+1);
        for(int j=1;j<=n;j++)
            b[j]=a[j]%mod;
        sort(b+1,b+1+n);
        int s=0;
        for(int j=1;j<=n;j++){
            int l,r;
            l=lower_bound(b+1,b+1+n,(1<<i)-b[j])-b;
            r=lower_bound(b+1,b+1+n,(1<<(i+1))-b[j])-b-1;
            s+=r-l+1;
            l=lower_bound(b+1,b+1+n,(1<<i)+(1<<(i+1))-b[j])-b;
            r=lower_bound(b+1,b+1+n,(1<<(i+2))-b[j])-b-1;
            s+=r-l+1;
            if((b[j]+b[j])&(1<<i)) s--;
        }
        if((s/2)&1) ans+=1<<i;
    }
    cout<<ans;
	return 0;
}

本文标签： DivBasedcodeforcesMoscowInformatics

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