CSES - Aalto Competitive Programming 2024 - wk4 - Mon - Results
Submission details
Task:Xor sum
Sender:paulschulte
Submission time:2024-09-23 16:21:12 +0300
Language:C++ (C++17)
Status:READY
Result:ACCEPTED
Test results
testverdicttime
#1ACCEPTED0.00 sdetails
#2ACCEPTED0.40 sdetails

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 val
using 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 s
for (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 here
int 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