1847:A — The Man who became a God
- Author — PoPularPlusPlus
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define pb(e) push_back(e)
#define sv(a) sort(a.begin(),a.end())
#define sa(a,n) sort(a,a+n)
#define mp(a,b) make_pair(a,b)
#define all(x) x.begin(),x.end()
void solve(){
int n , k;
cin >> n >> k;
ll arr[n];
for(int i = 0; i < n; i++)cin >> arr[i];
vector<ll> v;
ll sum = 0;
for(int i = 1; i < n; i++){
v.pb(abs(arr[i] - arr[i-1]));
sum += v.back();
}
sort(all(v));
for(int groups = 1; groups < k; groups++){
sum -= v.back();
v.pop_back();
}
cout << sum << '\n';
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;cin >> t;while(t--)
solve();
return 0;
}
- Author — PoPularPlusPlus
#include <iostream>
#include <vector>
using namespace std;
#define ll long long
#define ull unsigned long long
#define pb(e) push_back(e)
#define sv(a) sort(a.begin(),a.end())
#define sa(a,n) sort(a,a+n)
#define mp(a,b) make_pair(a,b)
#define all(x) x.begin(),x.end()
void solve(){
int n;
cin >> n;
int arr[n];
for(int i = 0; i < n; i++)cin >> arr[i];
int cur = arr[0];
int part = 1;
for(int i = 0; i < n; i++){
cur &= arr[i];
if(cur == 0){
if(i == n-1)break;
part++;
cur = arr[i + 1];
}
}
if(cur != 0)part--;
part = max(part,1);
cout << part << '\n';
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;cin >> t;while(t--)
solve();
return 0;
}
1847:C- Vampiric Powers, anyone?
- Author — PoPularPlusPlus
At the end of the array, you can only achieve xor of any subarray of the original array.
Lets denote $$$f(u,v) =$$$ xor of all $$$a_i$$$ such that $$$min(u,v) \leq i < max(u,v)$$$. In the first operation you add $$$f(n,i)$$$. I.e. $$$[u_1,v_1)=[n,i)$$$. It can be proven that $$$f(u_k,v_k) = f(v_{k-1},v_k)$$$ in the $$$k$$$-th operation which is a range.
Suppose we have taken $$$k$$$ ranges that already satisfy this property. Now, I add a new $$$k+1$$$-th range. So, first I need to take the $$$k$$$-th range $$$f(u_k,v_k)$$$. Now I'm xoring it with the range $$$f(u_{k - 1}, v_{k - 1})$$$. As [ $$$u_k, v_k$$$) and [ $$$u_{k - 1}, v_{k - 1}$$$) share an endpoint, the result for these ranges will be a range.
For two ranges $$$f(x,y)$$$ and $$$f(y,z)$$$, if the two ranges do not intersect, the result will be the sum of the two ranges $$$f(x,z)$$$. If the two ranges intersect, then the intersections will be cancelled out, and the result will be the difference $$$f(x,z)$$$.
Now, we maintain a boolean array $$$b$$$ where $$$b_i$$$ is $$$1$$$ if there is some $$$j$$$ such that $$$a_1 \oplus a_2 \oplus \cdots \oplus a_j = i$$$. Initially, $$$b$$$ is all $$$0$$$. We loop $$$j$$$ from $$$1$$$ to $$$n$$$ and check for each $$$k$$$ if $$$b_k=1$$$. If it is, then there is some position $$$p < j$$$ such that $$$a_1 \oplus a_2 \oplus \cdots \oplus a_p = k$$$. If we take xor of range from $$$(p,j]$$$, then it will be $$$k \oplus a_1 \oplus a_2 \oplus \cdots \oplus a_j$$$ (as $$$a_1 \oplus a_2 \oplus \cdots \oplus a_p$$$ gets cancelled). This $$$a_1 \oplus a_2 \oplus \cdots \oplus a_j$$$ can be stored as we loop ahead. We are looping all possible prefix xors and not all prefix positions because $$$n$$$ is large.
Time Complexity — $$$O(n \cdot 2^8)$$$.
#include <bits/stdc++.h>
using namespace std;
int main() {
cin.tie(0)->sync_with_stdio(0);
int ntest;
cin >> ntest;
while (ntest--) {
int n;
cin >> n;
vector<int> a(n);
for (auto &i : a)
cin >> i;
int const max_value = 1 << 8;
vector<char> has_pref(max_value);
has_pref[0] = true;
int cur_xor = 0;
int ans = 0;
for (auto i : a) {
cur_xor ^= i;
for (int pref = 0; pref < max_value; ++pref) {
if (has_pref[pref]) {
ans = max(ans, pref ^ cur_xor);
}
}
has_pref[cur_xor] = true;
}
cout << ans << '\n';
}
}
- Author — PoPularPlusPlus
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define pb(e) push_back(e)
#define sv(a) sort(a.begin(),a.end())
#define sa(a,n) sort(a,a+n)
#define mp(a,b) make_pair(a,b)
#define vf first
#define vs second
#define all(x) x.begin(),x.end()
void solve(){
int n , m , q;
cin >> n >> m >> q;
string st;
cin >> st;
vector<pair<int,int>> ranges;
for(int i = 0; i < m; i++){
int l , r;
cin >> l >> r;
l--;
r--;
ranges.pb(mp(l,r));
}
set<int> s;
for(int i = 0; i < n; i++)s.insert(i);
vector<int> v;
int pos_in_v[n];
memset(pos_in_v,-1,sizeof pos_in_v);
for(int i = 0; i < m; i++){
auto it = s.lower_bound(ranges[i].vf);
vector<int> toerase;
while(it != s.end() && (*it) <= ranges[i].vs){
toerase.pb((*it));
v.pb(toerase.back());
pos_in_v[toerase.back()] = v.size()-1;
it++;
}
while(toerase.size()){
s.erase(toerase.back());
toerase.pop_back();
}
}
int cnt = 0;
for(int i = 0; i < n; i++){
if(st[i] == '1')cnt++;
}
int ans = 0;
for(int i = 0; i < min(cnt , (int)v.size()); i++){
if(st[v[i]] == '0')ans++;
}
while(q--){
int pos;
cin >> pos;
pos--;
if(pos_in_v[pos] != -1 && pos_in_v[pos] < cnt){
if(st[pos] == '0'){
ans--;
}
else ans++;
}
if(st[pos] == '0'){
st[pos] = '1';
cnt++;
if(cnt <= v.size() && st[v[cnt-1]] == '0')ans++;
}
else {
st[pos] = '0';
if(cnt > 0 && cnt <= v.size() && st[v[cnt-1]] == '0')ans--;
cnt--;
}
cout << ans << '\n';
}
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
//int t;cin >> t;while(t--)
solve();
return 0;
}
- Author — StArChAn
- Idea — astoria
- Prepared by — PoPularPlusPlus
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define pb(e) push_back(e)
#define sv(a) sort(a.begin(),a.end())
#define sa(a,n) sort(a,a+n)
#define mp(a,b) make_pair(a,b)
#define vf first
#define vs second
#define all(x) x.begin(),x.end()
const int B = 31;
struct item {
ll pos , bit, idx;
};
item make(ll pos , ll bit , ll idx){
item res;
res.pos = pos;
res.bit = bit;
res.idx = idx;
return res;
}
void solve(){
int n;
cin >> n;
int qu;
cin >> qu;
int arr[n];
for(int i = 0; i < n; i++)cin >> arr[i];
if(n == 1){
while(qu--){
int x;
cin >> x;
if(x < arr[0])cout << 1 << '\n';
else cout << -1 << '\n';
}
return;
}
vector<pair<ll,ll>> v;
queue<item> q;
bool vis[n][B];
memset(vis,0,sizeof vis);
for(int i = 0; i < n; i++){
for(int j = 0; j < B; j++){
if((arr[i] & (1 << j)) > 0){
vis[i][j] = 1;
}
}
v.pb(mp(i , arr[i]));
}
for(int i = 0; i < n - 1; i++){
v.pb(mp(n+i,arr[i]|arr[i+1]));
}
for(int j = 0; j < B; j++){
if(vis[n-1][j])q.push(make(n+n-1,j,0));
}
for(int i = 0; i < n - 1; i++){
for(int j = 0; j < B; j++){
if(vis[i][j])q.push(make(n+n+i,j,(i+1)%(n-1)));
}
}
int val[n-1];
for(int i = 0; i < n-1; i++)val[i] = arr[i]|arr[i+1];
while(q.size()){
item it = q.front();
q.pop();
if((val[it.idx] & (1 << it.bit)) > 0){
continue;
}
val[it.idx] += (1 << it.bit);
if(v.back().vf == it.pos){
v.pop_back();
}
v.pb(mp(it.pos , val[it.idx]));
q.push(make(it.pos + n , it.bit , (it.idx + 1)%(n-1)));
}
// first part gets over where v[i].first contains position in ascending order and v[i].second contains the value to what has become on that position.
for(int i = 1; i < v.size(); i++){
v[i].vs = max(v[i].vs , v[i-1].vs);
}
while(qu--){
int x;
cin >> x;
ll l = 0 , r = v.size()-1;
ll res = -1;
while(l <= r){
int mid = (l + r)/2;
if(v[mid].vs <= x){
l = mid + 1;
}
else {
r = mid - 1;
res = v[mid].vf+1;
}
}
cout << res << '\n';
}
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;cin >> t;while(t--)
solve();
return 0;
}
