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); } }