原题链接

题解

1.如果 \(x \oplus y \gt x\) ，则 \(y\) 的最高位对应的 \(x\) 一定是 \(0\)
2.$f(x,y)\oplus f(y,z) \gt f(x,z) $ 等价于 \(f(x,z) \oplus a_y \gt f(x,z)\)
3.\(x \in [1,y]\) 则 \(x-1 \in [0,y-1]\)

code

#include<bits/stdc++.h>
using namespace std;

#define ll long long

ll a[100005]={0}, pre[100005]={0};
ll cntleft[100005][35][2]={0}, cntright[100005][35][2]={0};

inline void read(ll &x) {
	x = 0;
	ll flag = 1;
	char c = getchar();
    while(c < '0' || c > '9'){
        if(c == '-')flag = -1;
        c = getchar();
    }
	while(c >= '0' && c <= '9') {
		x = (x << 3) + (x << 1) + (c ^ 48);
		c = getchar();
	}
	x *= flag;
}

inline void write(ll x)
{
    if(x < 0){
    	putchar('-');
		x = -x;
	}
    if(x > 9)
		write(x / 10);
    putchar(x % 10 + '0');
}

int main()
{
    ll t;
    read(t);
    while(t--)
    {
        ll n;
        read(n);
        for(ll i=1; i<=n; i++)
        {
            read(a[i]);
            pre[i]=pre[i-1]^a[i];
        }

        for(ll k=0; k<=30; k++) cntleft[0][k][0]=1;
        for(ll i=1; i<=n; i++)
        {
            for(ll k=0; k<=30; k++)
            {
                ll add=((pre[i]>>k)&1);
                cntleft[i][k][add]=cntleft[i-1][k][add]+1;
                cntleft[i][k][1-add]=cntleft[i-1][k][1-add];
            }
        }

        for(ll k=0; k<=30; k++) cntright[n+1][k][1]=cntright[n+1][k][0]=0;
        for(ll i=n; i>=1; i--)
        {
            for(ll k=0; k<=30; k++)
            {
                ll add=((pre[i]>>k)&1);
                cntright[i][k][add]=cntright[i+1][k][add]+1;
                cntright[i][k][1-add]=cntright[i+1][k][1-add];
            }
        }

        ll ans=0;
        for(ll i=1; i<=n; i++)
        {
            ll height=log2(a[i]);
            ans+=cntleft[i-1][height][0]*cntright[i][height][0];
            ans+=cntleft[i-1][height][1]*cntright[i][height][1];
        }
        write(ans);
        putchar('\n');
    }
    return 0;
}