BZOJ4888 Tjoi2017异或和（树状数组）

　　化为前缀和相减。考虑每一位的贡献。则需要快速查询之前有几个数和当前数的差在第k位上为1。显然其与更高位是无关的。于是用BIT维护后k位的数的出现次数，瞎算一算即可。

// luogu-judger-enable-o2
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 100010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
int n,a[N],tree[][<<|],ans;
void ins(int p,int k){k++;while (k<=(<<)) tree[p][k]^=,k+=k&-k;}
int query(int p,int k){k++;int s=;while (k) s^=tree[p][k],k-=k&-k;return s;}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4888.in","r",stdin);
    freopen("bzoj4888.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read();
    for (int i=;i<=n;i++) a[i]=a[i-]+read();
    for (int i=;i<;i++) ins(i,);
    for (int i=;i<=n;i++)
        for (int j=;j<;j++)
        {
            int inf=(<<j+)-,x=a[i]&inf,r=x^(<<j),l=x+&inf;
            if ((l<=r?query(j,r)-query(j,l-):query(j,r)+query(j,inf)-query(j,l-))+&) ans^=<<j;
            ins(j,x);
        }
    cout<<ans;
    return ;
}