E. Distance Learning Courses in MAC：

题目大意：

思路解析：

// 对于这种二进制多个数计算答案，我们应该灵敏的想到是否可以通过枚举二进制位来计算答案。

就是对每一个查询找出或和的最大值，那我们想xi 和 yi中哪些位一定会出现在答案中，假设为25 和 31，他们两转为二进制为 (11001) 和 （11111）我们可以想到24一定会进入答案，如果它不是答案的一部分，那无论怎么选都无法满足选择的数大于等于x。那我们这样就可以对[l,r]的答案进行简单计算（这里利用线段树或者树状数组的区间查询即可），那后续剩下的答案怎么办。后续 x 和 y剩下的二进制位为 （00001） 和 （0111）我们发现对r的剩余无论选择哪一位都可以满足大于等于x，那我们可以对于这些剩下的数做一个前缀和的处理，如果有我们一定会把它假如答案。

那么如果假如剩下 有00100这一位，之前粗略的答案中也有00100这一位可以发现我们可以将其转化为00111，后面就可以不用枚举了。

代码实现：

import java.io.*;
import java.math.BigInteger;
import java.util.*;public class Main {static int inf = (int) 1e9;static int mod = 998244353;public static void main(String[] args) throws IOException {int t = f.nextInt();while (t > 0) {solve();t--;}w.flush();w.close();br.close();}static int n;static int[] l;static int[] r;static int[] v;static int maxN = (int) 2e5 + 5;static int[] t = new int[maxN * 2];public static void solve() {n = f.nextInt();l = new int[n + 1];r = new int[n + 1];v = new int[n + 1];for (int i = 1; i <= n; i++) {l[i] = f.nextInt();r[i] = f.nextInt();}fix();int[][] bits = new int[31][n + 1];for (int i = 1; i <= n; i++) {update(i, v[i]);for (int j = 30; j >= 0; j--) {bits[j][i] = bits[j][i-1];if (((r[i] >> j) & 1) == 1) bits[j][i]++;}}int q = f.nextInt();for (int i = 0; i < q; i++) {int x = f.nextInt();int y = f.nextInt();int ans = sum(x, y);for (int j = 30; j >= 0; j--) {int cnt = bits[j][y] - bits[j][x-1] + ((ans >> j) & 1);if (cnt > 1) {ans |= (2 << j) - 1;break;} else if (cnt == 1) {ans |= (1 << j);}}w.print(ans + " ");}w.println();for (int i = 0; i <= n; i++) {t[i] = 0;}}//    public static void fix(){
//        for (int i = 1; i <= n; i++) {
//            if (l[i] == r[i]){
//                v[i] = r[i];
//                l[i] = r[i] = 0;
//                continue;
//            }
//            int pref = (1 << (lg(l[i] ^ r[i]) + 1)) - 1;
//            v[i] = r[i] - (r[i] & pref);
//            r[i] = r[i] & pref;
//            l[i] = l[i] & pref;
//        }
//    }public static void fix() {for (int i = 1; i <= n; ++i) {if (l[i] == r[i]) {v[i] = l[i];l[i] = r[i] = 0;continue;}int pref = (1 << (lg(l[i] ^ r[i]) + 1)) - 1;v[i] = r[i] - (r[i] & pref);r[i] &= pref;l[i] &= pref;}}public static void update(int x, int val) {for (int i = x; i <= n; i += lowbit(i)) {t[i] |= val;}}public static int lg(int x) {for (int i = 30; i >= 0; i--) {if (((x >> i) & 1) == 1) return i;}return 0;}public static int sum(int l, int r) {int res = 0;while (l <= r) {res |= v[r];r--;while (r - lowbit(r) >= l) {res |= t[r];r -= lowbit(r);}}return res;}public static int lowbit(int x) {return x & -x;}static PrintWriter w = new PrintWriter(new OutputStreamWriter(System.out));static Input f = new Input(System.in);static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));static class Input {public BufferedReader reader;public StringTokenizer tokenizer;public Input(InputStream stream) {reader = new BufferedReader(new InputStreamReader(stream), 32768);tokenizer = null;}public String next() {while (tokenizer == null || !tokenizer.hasMoreTokens()) {try {tokenizer = new StringTokenizer(reader.readLine());} catch (IOException e) {throw new RuntimeException(e);}}return tokenizer.nextToken();}public String nextLine() {String str = null;try {str = reader.readLine();} catch (IOException e) {// TODO 自动生成的 catch 块e.printStackTrace();}return str;}public int nextInt() {return Integer.parseInt(next());}public long nextLong() {return Long.parseLong(next());}public Double nextDouble() {return Double.parseDouble(next());}public BigInteger nextBigInteger() {return new BigInteger(next());}}
}