Task: | Xor sum |
Sender: | paulschulte |
Submission time: | 2024-09-23 16:21:12 +0300 |
Language: | C++ (C++17) |
Status: | READY |
Result: | ACCEPTED |
test | verdict | time | |
---|---|---|---|
#1 | ACCEPTED | 0.00 s | details |
#2 | ACCEPTED | 0.40 s | details |
Code
// ~/.vim/cpp_template.cpp#include <bits/stdc++.h>#include <iostream>#include <vector>#include <algorithm>#include <string>#define REP(i,a,b) for (int i = a; i < b; i++)#define ll long long// Type Aliases for 1D and 2D vectors with initialization#define vi(n, val) vector<int>(n, val) // 1D vector of ints with size n, initialized to val#define vll(n, val) vector<long long>(n, val) // 1D vector of long longs with size n, initialized to val#define ll long long#define vvi(n, m, val) vector<vector<int>>(n, vector<int>(m, val)) // 2D vector of ints (n x m), initialized to val#define vvll(n, m, val) vector<vector<long long>>(n, vector<long long>(m, val)) // 2D vector of long longs (n x m), initialized to valusing namespace std;template <typename T>void pV(const std::vector<T>& vec, const std::string& label = "Vector") {std::cout << label << ": [ ";for (const auto& elem : vec) {std::cout << elem << " ";}std::cout << "]" << std::endl;}using namespace std;void dfs(int s, vector<bool> *visited, vector<int> (*adj)[]) {if ((*visited)[s]) return;(*visited)[s] = true;// process node sfor (auto u: (*adj)[s]) {dfs(u, visited, adj);}}/*vector<int> adj[N];vector<bool> visited(N, false);int u, v;for(int i = 0; i < M;i++){cin >> u >> v;u--;v--;adj[u].push_back(v);adj[v].push_back(u);}*/int qxor(int a, int b, ll n, vector<int>* tree) {a += n; b += n;int s = 0;while (a <= b) {//cout << a << ' ' << b << endl;if (a%2 == 1) s = (*tree)[a++] ^ s;if (b%2 == 0) s = (*tree)[b--] ^ s;a /= 2; b /= 2;}return s;}void change(int k, int u, ll n, vector<int>* tree) {k += n;(*tree)[k] = u;for (k /= 2; k >= 1; k /= 2) {(*tree)[k] = min((*tree)[2*k], (*tree)[2*k+1]);}}int main() {ios::sync_with_stdio(0);cin.tie(0);// Your code starts hereint n, q;cin >> n >> q;ll n_2 = pow(2, ceil(log(n)/log(2)));vector<int> v = vi(2*n_2+1, 1e9);for(ll i = n_2; i < n_2+n; i++){cin >> v[i];}for(ll i = n_2-1; i > 0; i--){v[i] = v[2*i] ^ v[2*i+1];}REP(i, 0, q){int a, b;cin >> a >> b;++a;++b;if(a == b){cout << v[n_2-2+a] << endl;continue;}cout << qxor(a, b, n_2-2, &v) << endl;}return 0;}
Test details
Test 1
Verdict: ACCEPTED
input |
---|
8 36 7 6 4 6 2 9 4 8 1 1 1 2 1 3 ... |
correct output |
---|
7 1 5 3 1 ... |
user output |
---|
7 1 5 3 1 ... |
Test 2
Verdict: ACCEPTED
input |
---|
200000 200000 921726510 307633388 992247073 ... |
correct output |
---|
834756431 130379787 403037296 308618218 784778243 ... |
user output |
---|
834756431 130379787 403037296 308618218 784778243 ... Truncated |