XOR key
Problem Statement :
Xorq has invented an encryption algorithm which uses bitwise XOR operations extensively. This encryption algorithm uses a sequence of non-negative integers as its key. To implement this algorithm efficiently, Xorq needs to find maximum value of for given integers , and , such that, . Help Xorq implement this function. For example, , , and . We test each for all values of between and inclusive: Function Description Complete the xorKey function in the editor below. It should return an integer array where each value is the response to a query. xorKey has the following parameters: x: a list of integers queries: a two dimensional array where each element is an integer array that consists of for the query at indices and respectively. Input Format The first line contains an integer , the number of test cases. The first line of each test case contains two space-separated integers and , the size of the integer array and the number of queries against the test case. The next line contains space-separated integers . Each of next lines describes a query which consists of three integers and . Output Format For each query, print the maximum value for , such that, on a new line.
Solution :
Solution in C :
In C :
#define NDEBUG
#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
struct header {
int* head;
int count;
};
int find(const int* base, int count, int p, int q)
{
int left = 0;
int right = count;
int mid;
int r;
while (left < right) {
mid = (left + right) / 2;
r = base[mid];
if (r > q) {
right = mid; // else r <= q
} else if (r < p) {
left = mid + 1; // else p <= r <= q
} else {
return r;
}
}
return -1;
}
int main(void)
{
int T, t;
int N, n;
int Q, r;
int a, p, q;
struct header* info;
int* table;
int i, j;
int ret;
int bit;
int next_bit;
int low;
int hlow;
int parent;
int child_low;
int child_high;
int low_ind;
int high_ind;
info = (struct header*)malloc(sizeof(struct header) * 65535);
assert(info != NULL);
table = (int*)malloc(sizeof(int) * 1600000);
assert(table != NULL);
ret = scanf("%d", &T);
assert(ret == 1);
assert(1 <= T && T <= 6);
for (t = 0; t < T; ++t) {
ret = scanf("%d%d", &N, &Q);
assert(ret == 2);
assert(1 <= N && N <= 100000);
assert(1 <= Q && Q <= 50000);
info[0].head = table;
info[0].count = N;
low = 0;
for (n = 0; n < N; ++n) {
ret = scanf("%d", table + n);
assert(ret == 1);
assert(0 <= table[n] && table[n] < 0x8000);
if ((table[n] & 0x4000) == 0)
++low;
}
info[1].count = low;
info[2].count = N - low;
// first iteration
bit = 0x4000;
next_bit = bit >> 1;
info[1].head = table + N;
info[2].head = info[1].head + low;
low = 0;
hlow = 0;
low_ind = 0;
high_ind = 0;
for (i = 0; i < N; ++i) {
if ((table[i] & bit) == 0) {
info[1].head[low_ind++] = i;
if ((table[i] & next_bit) == 0)
++low;
} else {
info[2].head[high_ind++] = i;
if ((table[i] & next_bit) == 0)
++hlow;
}
}
assert(low_ind == info[1].count);
assert(high_ind == info[2].count);
info[3].count = low;
info[4].count = low_ind - low;
info[5].count = hlow;
info[6].count = high_ind - hlow;
parent = 0;
for (bit = next_bit; bit > 0; bit = next_bit) {
next_bit = bit >> 1;
for (i = 0; i < 0x4000; i += bit) {
++parent;
child_low = parent + parent + 1;
child_high = child_low + 1;
info[child_low].head = info[child_low - 1].head + info[child_low - 1].count;
info[child_high].head = info[child_low].head + info[child_low].count;
low = 0;
hlow = 0;
low_ind = 0;
high_ind = 0;
for (j = 0; j < info[parent].count; ++j) {
if ((table[info[parent].head[j]] & bit) == 0) {
info[child_low].head[low_ind++] = info[parent].head[j];
if ((table[info[parent].head[j]] & next_bit) == 0)
++low;
} else {
info[child_high].head[high_ind++] = info[parent].head[j];
if ((table[info[parent].head[j]] & next_bit) == 0)
++hlow;
}
}
assert(low_ind == info[child_low].count);
assert(high_ind == info[child_high].count);
if (next_bit > 0) {
j = child_low + child_low + 1;
info[j++].count = low;
info[j++].count = low_ind - low;
info[j++].count = hlow;
info[j++].count = high_ind - hlow;
}
}
}
for (r = 0; r < Q; ++r) {
ret = scanf("%d%d%d", &a, &p, &q);
assert(ret == 3);
assert(0 <= a && a < 0x8000);
assert(1 <= p && p <= q && q <= N);
--p;
--q;
i = 0;
for (bit = 0x4000; bit > 0; bit >>= 1) {
if ((a & bit) == 0) {
i = i + i + 2;
if (find(info[i].head, info[i].count, p, q) < 0)
--i;
} else {
i = i + i + 1;
if (find(info[i].head, info[i].count, p, q) < 0)
++i;
}
assert(find(info[i].head, info[i].count, p, q) >= 0);
}
j = find(info[i].head, info[i].count, p, q);
assert(j >= 0);
printf("%d\n", a ^ table[j]);
}
}
free(table);
free(info);
return 0;
}
Solution in C++ :
In C ++ :
#include <iostream>
#include <cstdio>
#include <cstdio>
#include <cstring>
#include <cassert>
#include <algorithm>
#include <ctime>
using namespace std;
const int MB = 14;
const int N = 100000;
int arr[N], t, n, q, a, b, c;
int sorted_arrs[N * (MB + 2)];
struct XorqNode {
int child[2], idx, len;
} nodes[512 + N * (MB - 6)]; //
int arr_cnt, node_cnt;
int split(int idx, int len, int bm, int r) {
int cnt = 0;
for(int i = 0; i < len; i ++) {
if((bm & arr[sorted_arrs[i + idx]]) == r) {
sorted_arrs[arr_cnt++] = sorted_arrs[i + idx];
cnt ++;
}
}
return cnt;
}
void build(int root, int idx, int len, int bit = MB) {
nodes[root].idx = idx;
nodes[root].len = len;
if(bit == -1) return;
int bm = (1 << bit);
int sidx1 = arr_cnt;
int left_cnt = split(idx, len, bm, 0);
int sidx2 = arr_cnt;
int right_cnt = split(idx, len, bm, bm);
assert(left_cnt + right_cnt == len);
if(left_cnt) {
nodes[root].child[0] = node_cnt;
build(node_cnt ++, sidx1, left_cnt, bit - 1);
} else {
nodes[root].child[0] = -1;
}
if(right_cnt) {
nodes[root].child[1] = node_cnt;
build(node_cnt ++, sidx2, right_cnt, bit - 1);
} else {
nodes[root].child[1] = -1;
}
}
bool search(int root, int from, int to) {
int left = nodes[root].idx;
int right = left + nodes[root].len - 1;
int r = N;
while(left <= right) {
int mid = (left + right) / 2;
int val = sorted_arrs[mid];
if(val >= from) {
r = val;
right = mid - 1;
} else {
left = mid + 1;
}
}
return r <= to;
}
int query(int root, int n, int from, int to, int bit = MB) {
if(bit == -1) return 0;
int mybit = ((1 << bit) & n) ? 1 : 0;
if(nodes[root].child[1 - mybit] != -1 && search(nodes[root].child[1 - mybit], from, to)) {
return query(nodes[root].child[1 - mybit], n, from, to, bit - 1) + (1 << bit);
} else {
return query(nodes[root].child[mybit], n, from, to, bit - 1);
}
}
int query2(int n, int from, int to) {
int r = 0;
for(int i = from; i <= to; i ++) {
r = max(r, arr[i] ^ n);
}
return r;
}
int main() {
int cl = clock();
int err_cnt = 0;
for(scanf("%d", &t); t--; ) {
arr_cnt = node_cnt = 0;
scanf("%d %d", &n, &q);
for(int i = 0; i < n; i ++) {
scanf("%d", &arr[i]);
}
for(int i = 0; i < n; i ++) {
sorted_arrs[arr_cnt++] = i;
}
node_cnt = 1;
build(0, 0, n);
for(int i = 0; i < q; i ++) {
scanf("%d %d %d", &a, &b, &c);
int r1 = query(0, a, b - 1, c - 1);
cout << r1 << endl;
}
}
//cerr << (clock() - cl) * 0.001 << endl;
return 0;
}
Solution in Java :
In Java :
import java.util.*;
import java.io.*;
import java.math.BigInteger;
public class Solution {
private void solution() throws IOException {
int ts = in.nextInt();
while (ts-- > 0) {
int n = in.nextInt();
int q = in.nextInt();
int[] a = in.nextInts(n);
int size = 2 * Integer.highestOneBit(n);
int[][] sums = new int[size * 2][];
for (int i = 0; i < size; ++i) {
if (i < n) {
sums[i + size] = new int[] { a[i] };
} else {
sums[i + size] = new int[] {};
}
}
for (int i = size - 1; i > 0; --i) {
sums[i] = merge(sums[2 * i], sums[2 * i + 1]);
}
while (q-- > 0) {
int xor = in.nextInt();
int left = in.nextInt() - 1;
int right = in.nextInt() - 1;
int res = 0;
left += size;
right += size;
while (left <= right) {
if ((left & 1) != 0) {
res = Math.max(res, solve(sums[left++], xor));
}
if ((right & 1) == 0) {
res = Math.max(res, solve(sums[right--], xor));
}
left >>= 1;
right >>= 1;
}
out.println(res);
}
}
}
private int solve(int[] a, int xor) {
int left = 0;
int right = a.length - 1;
for (int bit = 14; bit >= 0; --bit) {
int middle = findMiddle(a, left, right, bit);
if (middle == left || middle > right) {
continue;
}
if (!get(xor, bit)) {
left = middle;
} else {
right = middle - 1;
}
}
if (a[left] != a[right]) {
throw null;
}
return a[left] ^ xor;
}
private int findMiddle(int[] a, int left, int right, int bit) {
while (left <= right) {
int middle = (left + right) >> 1;
if (!get(a[middle], bit)) {
left = middle + 1;
} else {
right = middle - 1;
}
}
return left;
}
private boolean get(int set, int bit) {
return (((set >> bit) & 1) != 0);
}
private int[] merge(int[] a, int[] b) {
int[] res = new int[a.length + b.length];
int i = 0;
int j = 0;
while (i < a.length && j < b.length) {
if (a[i] < b[j]) {
res[i + j] = a[i];
++i;
} else {
res[i + j] = b[j];
++j;
}
}
while (i < a.length) {
res[i + j] = a[i];
++i;
}
while (j < b.length) {
res[i + j] = b[j];
++j;
}
return res;
}
public void run() {
try {
solution();
in.reader.close();
out.close();
} catch (Throwable e) {
e.printStackTrace();
System.exit(1);
}
}
private void debug(Object... objects) {
System.out.println(Arrays.toString(objects));
}
private static class Scanner {
private BufferedReader reader;
private StringTokenizer tokenizer;
public Scanner(Reader reader) {
this.reader = new BufferedReader(reader);
this.tokenizer = new StringTokenizer("");
}
public boolean hasNext() throws IOException {
while (!tokenizer.hasMoreTokens()) {
String line = reader.readLine();
if (line == null) {
return false;
}
tokenizer = new StringTokenizer(line);
}
return true;
}
public String next() throws IOException {
hasNext();
return tokenizer.nextToken();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public double nextDouble() throws IOException {
return Double.parseDouble(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public String nextLine() throws IOException {
tokenizer = new StringTokenizer("");
return reader.readLine();
}
public int[] nextInts(int n) throws IOException {
int[] res = new int[n];
for (int i = 0; i < n; ++i) {
res[i] = nextInt();
}
return res;
}
public long[] nextLongs(int n) throws IOException {
long[] res = new long[n];
for (int i = 0; i < n; ++i) {
res[i] = nextLong();
}
return res;
}
public double[] nextDoubles(int n) throws IOException {
double[] res = new double[n];
for (int i = 0; i < n; ++i) {
res[i] = nextDouble();
}
return res;
}
public String[] nextStrings(int n) throws IOException {
String[] res = new String[n];
for (int i = 0; i < n; ++i) {
res[i] = next();
}
return res;
}
}
public static void main(String[] args) throws Exception {
new Solution().run();
}
private Scanner in = new Scanner(new InputStreamReader(System.in));
private PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
}
Solution in Python :
In Python3 :
from sys import stderr
MAXBITS = 15
def main():
ncases = int(input())
for case in range(ncases):
arrsize, nqueries = readints()
arr = readints()
assert arrsize == len(arr)
finder = XORkeyfinder(arr)
for query in range(nqueries):
a, start, stop = readints()
print(finder.findmax(a, start-1, stop))
def readints():
return [int(f) for f in input().split()]
class XORkeyfinder:
def __init__(self, arr):
self.tbl = []
self.arr = arr
for i in range(MAXBITS):
shift = MAXBITS -i - 1
prev = [None] * (2 << i)
row = [len(arr)] * len(arr)
chain = [None] * len(arr)
for loc, x in enumerate(arr):
x = x >> shift
p = prev[x ^ 1]
while p is not None:
row[p] = loc
p = chain[p]
# chain[p] = None
prev[x ^ 1] = None
chain[loc] = prev[x]
prev[x] = loc
self.tbl.append(row)
# print(' '.join('%2d' % i for i in row), file=stderr)
def findmax(self, a, start, stop, MAXBITS=MAXBITS):
arr, tbl = self.arr, self.tbl
comp = ~a
best, bestloc = arr[start], start
for i, row in enumerate(tbl):
test = 1 << (MAXBITS - i - 1)
if comp & test != best & test:
newloc = row[bestloc]
if newloc < stop:
best, bestloc = arr[newloc], newloc
return best ^ a
main()
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