Codeforces 1004F Sonya and Bitwise OR (线段树)

题目链接

https://codeforces.com/contest/1004/problem/F

题解

这种水题都不会做了怎么。。

考虑一个序列的前缀 \(\text{or}\) 值只会变化 \(O(\log W)\) 次，于是线段树维护每个区间的前缀和后缀 \(\text{or}\) 值即可。

时间复杂度 \(O(n\log n\log W)\).

代码

#include<bits/stdc++.h>
#define llong long long
#define mkpr make_pair
#define iter iterator
#define riter reversed_iterator
#define y1 Lorem_ipsum_dolor
#define pii pair<int,int>
using namespace std;

inline int read()
{
	int x = 0,f = 1; char ch = getchar();
	for(;!isdigit(ch);ch=getchar()) {if(ch==‘-‘) f = -1;}
	for(; isdigit(ch);ch=getchar()) {x = x*10+ch-48;}
	return x*f;
}

const int mxN = 1e5;
const int lgA = 20;
int a[mxN+3];
int n,q,X;

struct Data
{
	pii pre[lgA+3],suf[lgA+3]; llong sum; int pren,sufn;
} sgt[mxN*4+3];
Data operator +(Data ls,Data rs)
{
	Data ret; ret.sum = ls.sum+rs.sum; ret.pren = ret.sufn = 0;
	llong cur = 0ll;
	for(int i=1,j=rs.pren+1; i<=ls.sufn; i++)
	{
		while(j>1&&(rs.pre[j-1].first|ls.suf[i].first)>=X) {j--; cur += rs.pre[j].second;}
		ret.sum += 1ll*ls.suf[i].second*cur;
	}
	for(int i=1; i<ls.pren; i++) {ret.pre[++ret.pren] = ls.pre[i];}
	pii tmp = ls.pre[ls.pren];
	for(int i=1; i<=rs.pren; i++)
	{
		if((tmp.first|rs.pre[i].first)==tmp.first) {tmp.second += rs.pre[i].second;}
		else
		{
			ret.pre[++ret.pren] = tmp;
			tmp.first |= rs.pre[i].first; tmp.second = rs.pre[i].second;
		}
	}
	ret.pre[++ret.pren] = tmp;
	for(int i=1; i<rs.sufn; i++) {ret.suf[++ret.sufn] = rs.suf[i];}
	tmp = rs.suf[rs.sufn];
	for(int i=1; i<=ls.sufn; i++)
	{
		if((tmp.first|ls.suf[i].first)==tmp.first) {tmp.second += ls.suf[i].second;}
		else
		{
			ret.suf[++ret.sufn] = tmp;
			tmp.first |= ls.suf[i].first; tmp.second = ls.suf[i].second;
		}
	}
	ret.suf[++ret.sufn] = tmp;
	return ret;
}
void build(int u,int le,int ri)
{
	if(le==ri) {sgt[u].pren = sgt[u].sufn = 1,sgt[u].pre[1] = sgt[u].suf[1] = mkpr(a[le],1),sgt[u].sum = a[le]>=X?1:0; return;}
	int mid = (le+ri)>>1;
	build(u<<1,le,mid); build(u<<1|1,mid+1,ri);
	sgt[u] = sgt[u<<1]+sgt[u<<1|1];
}
void modify(int u,int le,int ri,int pos,int x)
{
	if(le==ri) {sgt[u].pre[1] = sgt[u].suf[1] = mkpr(x,1),sgt[u].sum = x>=X?1:0; return;}
	int mid = (le+ri)>>1;
	if(pos<=mid) modify(u<<1,le,mid,pos,x); else modify(u<<1|1,mid+1,ri,pos,x);
	sgt[u] = sgt[u<<1]+sgt[u<<1|1];
}
Data query(int u,int le,int ri,int lb,int rb)
{
	if(le>=lb&&ri<=rb) {return sgt[u];}
	int mid = (le+ri)>>1;
	if(rb<=mid) {return query(u<<1,le,mid,lb,rb);}
	else if(lb>mid) {return query(u<<1|1,mid+1,ri,lb,rb);}
	else {return query(u<<1,le,mid,lb,mid)+query(u<<1|1,mid+1,ri,mid+1,rb);}
}

int main()
{
	n = read(),q = read(),X = read();
	for(int i=1; i<=n; i++) a[i] = read();
	build(1,1,n);
	while(q--)
	{
		int opt = read();
		if(opt==1)
		{
			int u = read(),x = read();
			modify(1,1,n,u,x);
		}
		else
		{
			int l = read(),r = read();
			Data ans = query(1,1,n,l,r);
			printf("%I64d\n",ans.sum);
		}
	}
	return 0;
}

原文地址：https://www.cnblogs.com/suncongbo/p/12628829.html