luogu P4887 莫队二次离线

珂朵莉给了你一个序列$a$，每次查询给一个区间$[l,r]$

查询$l≤i<j≤r$,且$ai⊕aj$的二进制表示下有$k$个$1$的二元组$(i,j)$的个数。$⊕$是指按位异或。

直接暴力莫队的话复杂度是$O(nsqrt{m}inom{14}{7})$, 有一种做法是莫队二次离线

考虑莫队的过程, 假设当前维护的区间为$[ql,qr]$

若右端点移动到$r$, 那么答案的增量为

$sumlimits_{k=qr+1}^r f(k,[ql,k-1])=sumlimits_{k=qr+1}^r f(k,[1,k-1])-sumlimits_{k=qr+1}^rf(k,[1,ql-1])$

$f(k,[1,k-1])$可以预处理出来, $f(k,[1,ql-1])$可以离线以后扫描线.

总的复杂度就是$O(nsqrt{m}+ninom{14}{7})$

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



const int N = 1e5+10;
const int M = 1<<14;
int n, m, k, sqn, cnt;
int a[N], blo[N], f[N];
struct _ {
    int l,r,id;
	ll ans;
    bool operator < (const _ & rhs) const {
         return blo[l]^blo[rhs.l]?l<rhs.l:blo[l]&1?r<rhs.r:r>rhs.r;
    }
} q[N];
vector<_> g[N];

ll ans[N];
int vis[N],pre[N];

int main() {
	scanf("%d%d%d", &n, &m, &k);
	sqn = pow(n,0.55);
	REP(i,1,n) scanf("%d",a+i),blo[i]=i/sqn;
	REP(i,1,m) {
		scanf("%d%d",&q[i].l,&q[i].r);
		q[i].id = i;
	}
	if (k>14) {
		REP(i,1,m) puts("0");
		return 0;
	}
	REP(i,0,M-1) if (__builtin_popcount(i)==k) f[++cnt] = i;
	REP(i,1,n) {
		REP(j,1,cnt) pre[i]+=vis[f[j]^a[i]];
		++vis[a[i]];
	}
	sort(q+1,q+1+m);
	int ql=1, qr=0;
    REP(i,1,m) {
		int l = q[i].l, r = q[i].r;
		if (ql<l) g[qr].pb({ql,l-1,-i});
		while (ql<l) q[i].ans+=pre[ql++];
		if (ql>l) g[qr].pb({l,ql-1,i});
		while (ql>l) q[i].ans-=pre[--ql];
		if (qr<r) g[ql-1].pb({qr+1,r,-i});
		while (qr<r) q[i].ans+=pre[++qr];
		if (qr>r) g[ql-1].pb({r+1,qr,i});
		while (qr>r) q[i].ans-=pre[qr--];
    }
	REP(i,0,M-1) vis[i] = 0;
	REP(i,1,n) {
		REP(j,1,cnt) ++vis[a[i]^f[j]];
		for (auto &t:g[i]) {
			REP(j,t.l,t.r) {
				int ret = vis[a[j]];
				if (j<=i&&!k) --ret;
				if (t.id<0) q[-t.id].ans-=ret;
				else q[t.id].ans+=ret;
			}
		}
	}
	REP(i,1,m) q[i].ans+=q[i-1].ans;
	REP(i,1,m) ans[q[i].id] = q[i].ans;
	REP(i,1,m) printf("%lld
",ans[i]);
}