Ivan and Burgers CodeForces - 1100F (线性基)

时间：2024-08-13 21:05:50

大意: 给定n元素序列, m个询问$(l,r)$, 求$[l,r]$中选出任意数异或后的最大值

线性基沙茶题, 直接线段树暴力维护两个log还是能过的

#include <iostream>

#include <algorithm>

#include <cstdio>

#include <math.h>

#include <set>

#include <map>

#include <queue>

#include <string>

#include <string.h>

#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 '\n'

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;}

//head

const int N = 5e5+10;

int n, q, ql, qr;

int a[N];

struct _ {

	int a[22];

	void ins(int x) {

		REP(i,1,*a) x=min(x,a[i]^x);

		if (x) a[++*a]=x;

	}

	_ operator + (const _ &rhs) const {

		_ r;

		REP(i,0,*a) r.a[i]=a[i];

		REP(i,1,rhs.a[0]) r.ins(rhs.a[i]);

		return r;

	}

} tr[N<<2];

void build(int o, int l, int r) {

	if (l==r) return ({

		int t;

		scanf("%d", &t);

		tr[o].ins(t);

	});

	build(ls),build(rs);

	tr[o]=tr[lc]+tr[rc];

}

_ query(int o, int l, int r) {

	if (ql<=l&&r<=qr) return tr[o];

	if (mid>=qr) return query(ls);

	if (mid<ql) return query(rs);

	return query(ls)+query(rs);

}

int main() {

	scanf("%d", &n);

	build(1,1,n);

	scanf("%d", &q);

	REP(i,1,q) {

		scanf("%d%d",&ql,&qr);

		auto t = query(1,1,n);

		int ans = 0;

		REP(i,1,t.a[0]) ans=max(ans,ans^t.a[i]);

		printf("%d\n", ans);

	}

}

考虑一下一个log的做法, 对于每个基维护一个最后出现的位置, 贪心尽量让高位的位置最大就好了

#include <iostream>

#include <algorithm>

#include <cstdio>

#include <math.h>

#include <set>

#include <map>

#include <queue>

#include <string>

#include <string.h>

#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 '\n'

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;}

//head

const int N = 1e5+10;

int n, q;

int a[N][21], p[N][21];

int main() {

	scanf("%d", &n);

	REP(i,1,n) {

		int t;

		scanf("%d", &t);

		REP(j,0,20) a[i][j]=a[i-1][j],p[i][j]=p[i-1][j];

		int pos = i;

		PER(j,0,20) if (t>>j&1) {

			if (p[i][j]<pos) swap(a[i][j],t),swap(p[i][j],pos);

			if (!t) break;

			t ^= a[i][j];

		}

	}

	scanf("%d", &q);

	REP(i,1,q) {

		int l, r;

		scanf("%d%d", &l, &r);

		int ans = 0;

		PER(j,0,20) if (p[r][j]>=l) ans=max(ans,a[r][j]^ans);

		printf("%d\n", ans);

	}

}

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