Video Editorial
Idea: amul_agrawal
Problem Setting: menavlikar.rutvij JadeReaper rahulgoel amul_agrawal
Editorial: menavlikar.rutvij rahulgoel
Video Editorial: rahulgoel
Sum of two odd numbers is even and sum of two even numbers is also even.
If all consecutive pairs have even sum, can we generalize something about the sequence using the above hint?
#include <bits/stdc++.h>
using namespace std;
int main()
{
int t;
cin >> t;
while (t--)
{
int n;
cin >> n;
vector<int> a(n);
for (int i = 0; i < n; ++i)
cin >> a[i];
int num_odd = 0;
for (auto x : a)
if (x & 1)
num_odd++;
cout << min(num_odd, n - num_odd) << endl;
}
return 0;
}
1691B - Обмен обувью
Video Editorial
Idea: amul_agrawal
Problem Setting: rahulgoel menavlikar.rutvij JadeReaper
Editorial: menavlikar.rutvij rahulgoel
Video Editorial: rahulgoel
What happens to other people when a person receives a shoe greater his size?
#include <bits/stdc++.h>
using namespace std;
#define ll long long
typedef vector<ll> vll;
#define io \
ios_base::sync_with_stdio(false); \
cin.tie(NULL); \
cout.tie(NULL)
int main()
{
io;
ll tc;
cin >> tc;
while (tc--)
{
ll n;
cin >> n;
vll s(n), p(n);
for (ll i = 0; i < n; ++i)
cin >> s[i];
ll l = 0, r = 0;
bool ans = true;
for (ll i = 0; i < n; ++i)
p[i] = i + 1;
while (r < n)
{
while (r < n - 1 and s[r] == s[r + 1]) // get range [l,r] with equal values
++r;
if (l == r)
ans = false;
else
rotate(p.begin() + l, p.begin() + r, p.begin() + r + 1); // rotate right in range [l,r]
l = r + 1;
++r;
}
if (ans)
{
for (auto &x : p)
cout << x << " ";
cout << endl;
}
else
cout << -1 << endl;
}
return 0;
}
1691C - Сумма подстрок
Video Editorial
Idea: amul_agrawal
Problem Setting: rahulgoel amul_agrawal
Editorial: amul_agrawal rahulgoel
Video Editorial: rahulgoel
This is hint 1.
This is hint 2.
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin >> t;
while (t--) {
int n, k;
cin >> n >> k;
string s;
cin >> s;
int ones = 0, p1_first = n, p1_last = -1;
for (int p = 0; p < n; p++) {
if (s[p] != '1')
continue;
ones += 1;
if (p1_first == n)
p1_first = p;
p1_last = p;
}
int add = 0;
// moving the last one to last position
if (ones > 0 and (n - 1 - p1_last) <= k) {
k -= (n - 1 - p1_last);
add += 1;
ones -= 1;
}
// moving the first one to first position
if (ones > 0 and p1_first <= k) {
k -= (p1_first);
add += 10;
ones -= 1;
}
cout << 11 * ones + add << "\n";
}
return 0;
}
1691D - Максимум не меньше суммы
Video Editorial
Idea: amul_agrawal
Problem Setting: fangahawk amul_agrawal rahulgoel keyurchd_11
Editorial: fangahawk rahulgoel
Video Editorial: rahulgoel
This is hint 1.
This is hint 2.
#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
const ll ninf = -1e15;
vector<int> nextGreater(vector<ll>& arr, int n) {
stack<int> s;
vector<int> result(n, n);
for (int i = 0; i < n; i++) {
while (!s.empty() && arr[s.top()] < arr[i]) {
result[s.top()] = i;
s.pop();
}
s.push(i);
}
return result;
}
vector<int> prevGreater(vector<ll>& arr, int n) {
stack<int> s;
vector<int> result(n, -1);
for (int i = n - 1; i >= 0; i--) {
while (!s.empty() && arr[s.top()] < arr[i]) {
result[s.top()] = i;
s.pop();
}
s.push(i);
}
return result;
}
ll query(vector<ll> &tree, int node, int ns, int ne, int qs, int qe) {
if (qe < ns || qs > ne) return ninf;
if (qs <= ns && ne <= qe) return tree[node];
int mid = ns + (ne - ns) / 2;
ll leftQuery = query(tree, 2 * node, ns, mid, qs, qe);
ll rightQuery = query(tree, 2 * node + 1, mid + 1, ne, qs, qe);
return max(leftQuery, rightQuery);
}
int main() {
int t;
cin >> t;
while (t--) {
int n, _n;
cin >> n;
vector<ll> arr(n, 0);
for (auto& a : arr)
cin >> a;
// Round off n to next power of 2
_n = n;
while (__builtin_popcount(_n) != 1) _n++;
// Prefix sums
vector<ll> prefixSum(n, 0), suffixSum(n, 0);
prefixSum[0] = arr[0];
for (int i = 1; i < n; i++) {
prefixSum[i] = prefixSum[i - 1] + arr[i];
}
suffixSum[n - 1] = arr[n - 1];
for (int i = n - 2; i >= 0; i--) {
suffixSum[i] = suffixSum[i + 1] + arr[i];
}
// Two max-segtress, one on the prefix sums, one on the suffix sums
vector<ll> prefixTree(2 * _n, ninf), suffixTree(2 * _n, ninf);
for (int i = 0; i < n; i++) {
prefixTree[_n + i] = prefixSum[i];
suffixTree[_n + i] = suffixSum[i];
}
for (int i = _n - 1; i >= 1; i--) {
prefixTree[i] = max(prefixTree[2 * i], prefixTree[2 * i + 1]);
suffixTree[i] = max(suffixTree[2 * i], suffixTree[2 * i + 1]);
}
vector<int> ng = nextGreater(arr, n);
vector<int> pg = prevGreater(arr, n);
bool flag = true;
for (int i = 0; i < n; i++) {
ll rightMax = query(prefixTree, 1, 0, _n - 1, i + 1, ng[i] - 1) - prefixSum[i];
ll leftMax = query(suffixTree, 1, 0, _n - 1, pg[i] + 1, i - 1) - suffixSum[i];
if (max(leftMax, rightMax) > 0) {
flag = false;
break;
}
}
if (flag)
cout << "YES\n";
else
cout << "NO\n";
}
}
1691E - Число групп отрезков
Video Editorial
Idea: amul_agrawal
Problem Setting: akcube keyurchd_11 rahulgoel amul_agrawal
Editorial: akcube rahulgoel
Video Editorial: rahulgoel
This is hint 1.
This is hint 2.
#include <bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template <class T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define MOD 1000000007
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pb push_back
#define all(s) s.begin(), s.end()
#define sz(x) (int)(x).size()
#define fastio cin.tie(0) -> sync_with_stdio(0)
struct DSU{
vi dsu, szx;
DSU() = default;
DSU(int n) : dsu(n), szx(n, 1) {
for(int i=0; i<n; i++) dsu[i] = i;
}
int parent(int i){
if(dsu[i]==i) return i;
else return dsu[i] = parent(dsu[i]);
}
int size(int i) { return szx[parent(i)]; }
int operator[](int i){ return parent(i); }
int num_comps(){
int ct = 0;
for(int i=0; i<sz(dsu); i++) if(dsu[i] == i) ct++;
return ct;
}
void unify(int a, int b){
a = parent(a);
b = parent(b);
if(szx[a] < szx[b]) swap(a, b);
if(a!=b) dsu[b] = a, szx[a] += szx[b];
}
};
struct Point{
int t, p, pc, id;
bool close;
bool operator<(Point &other){
if(p == other.p) return close < other.close;
return p < other.p;
}
};
void solve(){
int n; cin>>n;
vector<Point> points;
for(int i=0; i<n; i++){
int t, l, r; cin>>t>>l>>r;
points.pb({t, l, r, i, false});
points.pb({t, r, l, i, true});
}
sort(all(points));
DSU dsu(n);
vector<set<pii>> open(2); // {r, id}
for(auto &p:points){
if(p.close) open[p.t].erase({p.p, p.id});
else{
open[p.t].insert({p.pc, p.id});
while(sz(open[p.t ^ 1]) > 1){
auto [r, id] = *open[p.t ^ 1].begin();
dsu.unify(p.id, id);
open[p.t ^ 1].erase({r, id});
}
if(sz(open[p.t ^ 1]) == 1) dsu.unify(p.id, open[p.t ^ 1].begin()->second);
}
}
cout<<dsu.num_comps()<<endl;
}
int main(){
fastio;
int t;
cin >> t;
while (t--) solve();
}
print("Hello World!")
1691F - Дерево и k-подмножества
Video Editorial
Idea: ltc.groverkss
Problem Setting: ltc.groverkss amul_agrawal rahulgoel
Editorial: rahulgoel
Video Editorial: rahulgoel
This is hint 1.
This is hint 2.
#include <iostream>
#include <vector>
using namespace std;
using ll = long long;
const ll MOD = 1e9 + 7;
struct Comb {
vector<ll> fac;
vector<ll> invfac;
ll n;
Comb(ll n)
{
this->n = n;
fac.resize(n + 1, 0);
invfac.resize(n + 1, 0);
fac[0] = 1;
for (ll i = 1; i <= n; i++)
fac[i] = (fac[i - 1] * i) % MOD;
invfac[n] = power(fac[n], MOD - 2);
for (ll i = n - 1; i >= 0; i--)
invfac[i] = (invfac[i + 1] * (i + 1)) % MOD;
}
static ll power(ll x, ll y)
{
ll ret = 1;
while (y) {
if (y & 1)
ret = (ret * x) % MOD;
y >>= 1;
x = (x * x) % MOD;
}
return ret;
}
ll nCr(ll n, ll r)
{
if (n < 0 or r < 0 or n < r)
return 0;
ll ans = (fac[n] * ((invfac[r] * invfac[n - r]) % MOD)) % MOD;
return ans;
}
};
vector<vector<int> > adj;
vector<int> sz;
vector<ll> cnt;
vector<ll> cntsz;
Comb C(2e5 + 5);
ll cur_ans = 0;
ll ans = 0;
void dfs1(int v, int p, int k)
{
sz[v] = 1;
ll sub = 0;
for (int u : adj[v]) {
if (u != p) {
dfs1(u, v, k);
sz[v] += sz[u];
sub = (sub + C.nCr(sz[u], k)) % MOD;
}
}
cnt[v] = (C.nCr(sz[v], k) - sub + MOD) % MOD;
cntsz[v] = (cnt[v] * sz[v]) % MOD;
cur_ans = (cur_ans + cntsz[v]) % MOD;
}
void dfs2(int v, int p, int k)
{
ans = (ans + cur_ans) % MOD;
for (int u : adj[v]) {
if (u != p) {
// store
int store_v_sz = sz[v];
ll store_v_cnt = cnt[v];
ll store_v_cntsz = cntsz[v];
int store_u_sz = sz[u];
ll store_u_cnt = cnt[u];
ll store_u_cntsz = cntsz[u];
ll store_cur_ans = cur_ans;
// recalculate size[v], size[u]
sz[v] -= sz[u];
sz[u] = sz.size();
// recalculate cnt[v]
cnt[v] = (cnt[v] - C.nCr(store_v_sz, k) + MOD) % MOD;
cnt[v] = (cnt[v] + C.nCr(sz[v], k)) % MOD;
cnt[v] = (cnt[v] + C.nCr(store_u_sz, k)) % MOD;
// recalculate cnt[u]
cnt[u] = (cnt[u] - C.nCr(store_u_sz, k) + MOD) % MOD;
cnt[u] = (cnt[u] + C.nCr(sz[u], k)) % MOD;
cnt[u] = (cnt[u] - C.nCr(sz[v], k) + MOD) % MOD;
// recalculate cntsz
cntsz[v] = (cnt[v] * sz[v]) % MOD;
cntsz[u] = (cnt[u] * sz[u]) % MOD;
// recalculate cur_ans
cur_ans = (cur_ans - store_v_cntsz - store_u_cntsz + MOD + MOD) % MOD;
cur_ans = (cur_ans + cntsz[v] + cntsz[u]) % MOD;
dfs2(u, v, k);
// restore
sz[v] = store_v_sz;
cnt[v] = store_v_cnt;
cntsz[v] = store_v_cntsz;
sz[u] = store_u_sz;
cnt[u] = store_u_cnt;
cntsz[u] = store_u_cntsz;
cur_ans = store_cur_ans;
}
}
}
int main()
{
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n, k;
cin >> n >> k;
adj.resize(n);
sz.resize(n);
cnt.resize(n);
cntsz.resize(n);
for (int i = 0; i < n - 1; ++i) {
int u, v;
cin >> u >> v;
--u, --v;
adj[u].push_back(v);
adj[v].push_back(u);
}
dfs1(0, 0, k);
dfs2(0, 0, k);
cout << ans << endl;
return 0;
}