题解
让异或值最大显然要按位贪心,然后我们还发现加上一个(x_i)的效果就是所有(a_i)整体向右偏移了,我们对于({a_i})开个主席树,支持查询一个区间中有多少个在([L,R])之间的数,查询时考虑一下(x_i)的影响就行了
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <cmath>
#include <ctime>
#include <queue>
#include <map>
#include <set>
using namespace std;
#define ull unsigned long long
#define pii pair<int, int>
#define uint unsigned int
#define mii map<int, int>
#define lbd lower_bound
#define ubd upper_bound
#define INF 0x3f3f3f3f
#define IINF 0x3f3f3f3f3f3f3f3fLL
#define vi vector<int>
#define ll long long
#define mp make_pair
#define pb push_back
#define re register
#define il inline
#define N 200000
#define M 100000
#define A 300000
#define LIM 17
int n, m;
int a[N+5];
int nid, root[N+5], sumv[50*N+5], ch[2][50*N+5];
void insert(int o, int &u, int l, int r, int x) {
u = ++nid;
sumv[u] = sumv[o]+1;
ch[0][u] = ch[0][o], ch[1][u] = ch[1][o];
if(l == r) return ;
int mid = (l+r)>>1;
if(x <= mid) insert(ch[0][o], ch[0][u], l, mid, x);
else insert(ch[1][o], ch[1][u], mid+1, r, x);
}
int query(int o, int u, int l, int r, int L, int R) {
if(L <= l && r <= R) return sumv[u]-sumv[o];
int mid = (l+r)>>1, ret = 0;
if(L <= mid) ret += query(ch[0][o], ch[0][u], l, mid, L, R);
if(R > mid) ret += query(ch[1][o], ch[1][u], mid+1, r, L, R);
return ret;
}
int Query(int b, int x, int l, int r) {
int ans = 0, t = 0;
for(int bit = LIM; bit >= 0; --bit) {
if((b>>bit)&1) {
if(query(root[l-1], root[r], 0, A, max(0, t-x), max(0, t+(1<<bit)-1-x))) {
ans |= (1<<bit);
}
else t |= (1<<bit);
}
else {
if(query(root[l-1], root[r], 0, A, max(0, t+(1<<bit)-x), max(0, t+(1<<(bit+1))-1-x))) {
t |= (1<<bit);
ans |= (1<<bit);
}
}
}
return ans;
}
int main() {
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; ++i) scanf("%d", &a[i]), insert(root[i-1], root[i], 0, A, a[i]);
for(int i = 1, b, x, l, r; i <= m; ++i) {
scanf("%d%d%d%d", &b, &x, &l, &r);
printf("%d
", Query(b, x, l, r));
}
return 0;
}