Let $s_{1 \dots n}$ be the string generated after applying the $m$ initial flip updates.
We can make $s$ by directly applying the flips in $O(nm)$ time. For each query, we can then use $O(2^n)$ time to enumerate all possible subsequences and take the lexicographically largest one.
This takes $O(nm + q 2^n)$ time, which runs in time for $n \leq 10, q \leq 1000$.
To make further progress on the problem, we consider the following greedy strategy:
Let $sz = r - l + 1$ and $z$ be the number of $0$'s in $s_{l \dots r}$. Let $t_{1 \dots k}$ be our final string.
Remember that comparing the lexicographical order of two strings $t_1$ and $t_2$ is equivalent to comparing the numbers they interpret in binary.
Thus, our solution that removes all the earliest $0$s (and $1$s when we run out of $0$s) is optimal.
For the sake of simplicity, we will analyze subtask 3 first.
We can loop over the elements in $s_{l \dots r}$ and directly apply our greedy strategy. If there are enough $1$'s, our answer is a string with all ones, otherwise, we remove the first $sz-k$ zeroes. Finally, we can interpret our answer as a binary number with the formula given.
Here is Alex Fan's C++ code:
using namespace std;
#include <iostream>
#include <set>
#include <algorithm>
#include <vector>
#include <map>
#define MAXN 200005
#define MOD 1000000007
int N, M, Q, a[MAXN];
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cin >> N >> M >> Q;
for(int i = 0;i < M;++i) {
int l, r;
cin >> l >> r;
l--;r--;
for(int i = l;i <= r;++i) {
a[i] ^= 1;
}
}
for(int i = 0;i < Q;++i) {
int l, r, k;
cin >> l >> r >> k;
l--;r--;
int owo = l;
int uwu = 0;
for(int i = l;i <= r;++i) {
uwu += a[i];
if(uwu + (r - i) >= k) owo = i;
}
uwu = 0;
for(int i = l;i <= owo;++i) uwu += a[i];
if(uwu >= k) {
int xwx = 0;
for(int i = 0;i < k;++i) {
xwx = (2 * xwx + 1) % MOD;
}
cout << xwx << endl;
continue;
}
int xwx = 0;
for(int i = 0;i < uwu;++i) {
xwx = (2 * xwx + 1) % MOD;
}
for(int i = owo + 1;i <= r;++i) {
xwx = (2 * xwx + a[i]) % MOD;
}
cout << xwx << endl;
}
return 0;
}
For subtask 2, we can observe that even though $n$ is large, we can describe it as $O(m)$ intervals of $0$'s or $1$'s. We can still apply our greedy but slightly modified to account for the larger intervals. This approach would require us to raise $2$ a large power modulo $10^9 +7$, which can be efficiently done using binary exponentiation. For more on binary exponentiation, refer to this USACO guide article. This takes $O(mq \log n)$ time, which runs in time for $m \leq 10$.
To improve past $O(n)$ or $O(m)$ per query, we must perform the greedy strategy more efficiently. We can binary search on the index $i$ of the $sz - k + 1$'th zero. This can be done with an auxiliary array $cnt$ that counts the number of $1$'s in the prefix of $1 \dots i-1$. Once we have found it, our string $t$ is $11 \dots 1$ for $i - l$ ones followed by $s_{i \dots r}$. Even though we cannot explicitly store $t$, we can still interpret it.
Suppose we precomputed prefix sums $\text{pre}_{1 \dots n}$ where $\text{pre}_{i} = \sum_{j = 1}^{i-1} s_j \cdot 2^{n - j - 1}$.
Then, $t$ will be intepretted as:
This can be evaluated in $O(\log n)$ or $O(1)$ time depending on how the powers of $2$ and $\frac{1}{2}$ mod $10^9 + 7$ are computed. Note that $\frac{1}{2} \equiv 500000004 \bmod 10^9 + 7$. There is an alternate approach using $pre_{i} = \sum_{j = 1}^{i} s_j \cdot 2^{i - j}$, which does not require inverse mod (division under mod).
This has complexity $O(n \log n + q \log n)$ or $O(n + q \log n)$ as we have to binary search on the index $i$.
Here is my code in C++:
//i dont code with uwus.... qaq
#include <iostream>
#include <algorithm>
#include <vector>
#include <array>
#include <map>
#define sz(x) ((int)(x.size()))
#define all(x) x.begin(), x.end()
using ll = long long;
const ll mod = 1e9 + 7;
const ll half = (mod + 1)/2;
const int MX = 2e5 + 10;
using namespace std;
int n, m, q;
ll pre[MX], cnt[MX], val[MX];
ll binpow(ll base, int p){ //as all the exponents are <= n, we can also precompute it
ll ans = 1;
for(int i = 0; (1ll << i) <= p; i++){
if(p&(1ll << i)){
(ans *= base) %= mod;
}
(base *= base) %= mod;
}
return ans;
}
int main(){
cin.tie(0) -> sync_with_stdio(0);
cin >> n >> m >> q ;
for(int i = 1; i<=m; i++){
int l, r;
cin >> l >> r;
val[l] ^= 1;
val[r+1] ^= 1;
}
for(int i = 1; i<=n+1; i++){
val[i] ^= val[i-1];
cnt[i] = cnt[i-1] + val[i-1];
pre[i] = pre[i-1] + val[i-1] * binpow(2, n - i + 1);
pre[i] %= mod;
}
while(q--){
int l, r, k;
cin >> l >> r >> k;
if(cnt[r+1] - cnt[l] >= k){
cout << binpow(2, k) - 1 << "\n";
continue;
}
int lo = 0, hi = n+1;
while(lo + 1 != hi){
int mid = (lo + hi)/2;
if(cnt[mid] - cnt[l] + (r + 1 - mid) >= k){
lo = mid;
}else{
hi = mid;
}
}
int i = lo;
ll front = binpow(2, k) - binpow(2, k - (cnt[i] - cnt[l]));
(front += mod) %= mod;
ll back = binpow(half, n - r) * (pre[r+1] - pre[i])%mod;
(back += mod) %= mod;
cout << (front + back)%mod << "\n";
}
}
Again, as $n$ is large, we can describe $s$ as $O(m)$ intervals of $0$'s and $1$'s. Additionally, with $q$ queries, we can determine $n' = O(m + q)$ relevant points for which to precompute information. Suppose the start points of these $0$/$1$ and query intervals in sorted order are $e_{1 \dots n'}$.
Imagine the two prefix sums array of length $n$, $cnt$ and $pre$, as the same as defined above. It is not practical to store both arrays explicitly, so we will only store the values at $e_1, e_2, \dots, e_{n'}$. Suppose we defined $cnt'_i = cnt_{e_i}$ and $pre'_{i} = pre_{e_i}$ (and $cnt'_0 = pre'_0 = 0$). We can compute $pre'$ efficiently by computing the powers of $2$ using binary exponentiation.
When we binary search, we find $j$ such that $e_j \leq i < e_{j+1}$. Then as all the elements in $s_{e_j \dots e_{j+1}-1}$ are either all $0$ or all $1$, we can directly solve for $i$ with casework.
When we finally interpret $t$, we can break the sum of $s_{i \dots r}$ into two parts: the prefix of $1$'s and the suffix of our subarray. The suffix of our subarray is given as $(1/2)^{n - r}(pre'_{r+1} - pre'_{i})$. Note that we cannot precompute the value of $pre'_{i}$ (as we don't know what $i$ is), but we can efficiently evaluate it on the fly. This again follows from having a precomputed value up until $e_j$ and knowing that the interval from $e_j$ to $i$ is either all $0$'s or all $1$'s.
This takes $O(\log m)$ time to binary search $i$, and $O(\log n)$ time to compute the powers of two per query. We also need $O((m + q) (\log (m + q) + \log n))$ time to precompute $cnt'$ and $pre'$. This gives our final complexity to be $O((m + q)(\log (m + q) + \log n))$. Other complexity with $\log^2$ complexities might be fast enough to pass.
Here is my C++ code:
//i dont code with uwus.... qaq
#include <iostream>
#include <algorithm>
#include <vector>
#include <array>
#include <map>
#define sz(x) ((int)(x.size()))
#define all(x) x.begin(), x.end()
using ll = long long;
const ll mod = 1e9 + 7;
const ll half = (mod + 1)/2;
using namespace std;
vector<int> e;
vector<ll> pre, cnt;
vector<int> val;
map<int, int> delta;
int n, m, q;
vector<array<int, 3>> qqs;
ll binpow(ll base, int p){
ll ans = 1;
for(int i = 0; (1ll << i) <= p; i++){
if(p&(1ll << i)){
(ans *= base) %= mod;
}
(base *= base) %= mod;
}
return ans;
}
#define gind(x) (lower_bound(all(e), x) - e.begin())
int main(){
cin.tie(0) -> sync_with_stdio(0);
cin >> n >> m >> q;
for(int i = 1; i<=m; i++){
int l, r;
cin >> l >> r;
delta[l] ^= 1;
delta[r+1] ^= 1;
e.push_back(l);
e.push_back(r+1);
}
for(int i = 1; i<=q; i++){
int l, r, k;
cin >> l >> r >> k;
qqs.push_back({l, r, k}); //no need to make queries offline, but makes impl easier
e.push_back(l);
e.push_back(r+1);
}
e.push_back(0);
e.push_back(n+1);
sort(all(e));
e.resize(unique(all(e)) - e.begin());
m = sz(e); //n'
val.resize(m);
pre.resize(m);
cnt.resize(m);
for(auto [i, v] : delta){
val[gind(i)] ^= v;
}
for(int i = 1; i<m; i++){
val[i] ^= val[i-1]; //value of s_e_i
cnt[i] += cnt[i-1]; //sum of [0, e_i)
pre[i] += pre[i-1]; //sum of [0, e_i)
cnt[i] += val[i-1] * (e[i] - e[i-1]);
pre[i] += val[i-1] * (binpow(2, n+1 - e[i-1]) - binpow(2, n+1 - e[i]));
pre[i] %= mod;
}
for(auto [l, r, k] : qqs){
int gl = gind(l);
int gr = gind(r+1);
if(cnt[gr] - cnt[gl] >= k){
cout << binpow(2, k) - 1 << "\n";
continue;
}
int cntl = cnt[gl];
int lo = 0, hi = m;
while(lo + 1 != hi){
int mid = (lo + hi)/2;
if(cnt[mid] - cntl + (r - e[mid]) >= k){
lo = mid;
}else{
hi = mid;
}
}
int i = cnt[lo] + r - cntl - k;
ll front = binpow(2, k) - binpow(2, k - (cnt[lo] - cntl));
ll prei = pre[lo] + val[lo] * (binpow(2, n+1 - e[lo]) - binpow(2, n+1 - i))%mod;
(prei += mod) %= mod;
ll back = binpow(half, n-r) * (pre[gr] - prei)%mod;
(front += mod) %= mod;
(back += mod) %= mod;
cout << (front + mod + back) % mod << "\n";
}
return 0;
}
Bonus: Can you solve this in $O(n + q)$ time?