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Progressive hints first, then the full explanation and implementation when you're ready to cash out.
Review status
AI-generated and still unreviewed. Double-check the details before internalizing them.
Hints
Open only as much as you need to keep the solve alive.
If a value works, then XOR by is a bijection from the interval to itself, not just into it. So include temporarily and think about the set of all valid masks as symmetries of .
The valid masks form a group under XOR: if and both preserve the interval, then also preserves it. So the answer must be for some dimension .
A mask with highest bit can only work if the interval has matching structure on the two sides of the boundary. Low bits act inside halves; the top bit swaps halves.
Shift the interval so its lowest value is . If preserves , then for every offset , both and differ by another offset in . The offset behavior is controlled by the binary carries of .
The answer is determined only by the length : the stabilizer has size exactly when is a power of two and the interval is one XOR-coset of that size; otherwise it recursively comes from the largest complete mirrored chunk. A bit-recursive interval DP computes this directly.
For a set , a mask is valid exactly when
The statement only says every frog must stay inside , but XOR by a fixed mask is a bijection. Since there are exactly frogs and exactly suitable pads, “inside” automatically means “onto”.
Now include for a moment. The valid masks form a group under XOR:
So the number of valid masks including zero is always a power of two. The final answer is that number minus .
Recursive Binary View
Look at the interval inside a block . Split the block into two halves:
and
A mask whose top bit is keeps numbers in their current half. A mask whose top bit is swaps the two halves. The lower bits then act recursively inside the normalized halves.
So we solve a slightly more general problem.
Let
be the set of all -bit masks that transform interval into interval , where both intervals live inside .
For the original answer, we need
Split interval into normalized lower and upper pieces , and do the same for : .
If the top bit of the mask is , then the same lower mask must satisfy both:
and
So those masks are
If the top bit of the mask is , then halves are swapped, so the lower mask must satisfy:
and
Then we add the top bit to those masks.
That gives a clean recursion.
How Do We Store Sets of Masks?
The sets of masks produced here are affine XOR spaces. That means each set looks like
where is one representative mask, and is a linear XOR basis.
So we store:
base,Then we need two operations:
For intervals, the union of the “top bit ” branch and the “top bit ” branch is still one affine space whenever it is non-empty. The reason is group structure: all masks transforming one fixed interval into another form either nothing or one coset of the stabilizer.
Intersection is just linear algebra over XOR. To intersect
and
we need solve
A standard XOR basis handles this in tiny time, since there are only about relevant bits.
Base Cases
If both intervals are empty, every -bit mask maps empty to empty, so the space is full dimension .
If exactly one interval is empty, no mask works.
If , both non-empty intervals are just {0}, so only mask works.
Complexity
, so we use at most bits. Each test case explores a small recursive tree and does XOR-basis operations on at most -bit numbers. This is easily fast enough in practice.
The last step subtracts one because is forbidden. Tiny detail, very easy to forget, very annoying if you do.
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
void setIO() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
}
struct Basis {
static const int B = 61;
unsigned long long a[B]{};
bool add(unsigned long long x) {
for (int i = B - 1; i >= 0; --i) {
if (!((x >> i) & 1ULL)) continue;
if (!a[i]) {
a[i] = x;
return true;
}
x ^= a[i];
}
return false;
}
vector<unsigned long long> vecs() const {
vector<unsigned long long> v;
for (int i = 0; i < B; ++i) if (a[i]) v.push_back(a[i]);
return v;
}
int rank() const {
int res = 0;
for (int i = 0; i < B; ++i) res += (a[i] != 0);
return res;
}
};
struct Space {
bool empty = true;
unsigned long long base = 0;
Basis basis;
};
Space none() { return Space{}; }
Space point(unsigned long long x) {
Space s;
s.empty = false;
s.base = x;
return s;
}
Space full_space(int k) {
Space s = point(0);
for (int i = 0; i < k; ++i) s.basis.add(1ULL << i);
return s;
}
Space normalize(Space s) {
if (s.empty) return s;
for (int i = Basis::B - 1; i >= 0; --i) {
if (((s.base >> i) & 1ULL) && s.basis.a[i]) s.base ^= s.basis.a[i];
}
return s;
}
Space intersect_space(const Space& A, const Space& B) {
if (A.empty || B.empty) return none();
vector<unsigned long long> u = A.basis.vecs();
vector<unsigned long long> v = B.basis.vecs();
int n = (int)u.size(), m = (int)v.size();
struct Row {
unsigned long long val = 0, mask = 0;
} row[61];
auto add = [&](unsigned long long val, unsigned long long mask) -> unsigned long long {
for (int bit = 60; bit >= 0; --bit) {
if (!((val >> bit) & 1ULL)) continue;
if (!row[bit].val) {
row[bit] = {val, mask};
return ULLONG_MAX;
}
val ^= row[bit].val;
mask ^= row[bit].mask;
}
return mask;
};
for (int i = 0; i < n; ++i) add(u[i], 1ULL << i);
for (int j = 0; j < m; ++j) add(v[j], 1ULL << (n + j));
unsigned long long need = A.base ^ B.base;
unsigned long long take = 0;
for (int bit = 60; bit >= 0; --bit) {
if (!((need >> bit) & 1ULL)) continue;
if (!row[bit].val) return none();
need ^= row[bit].val;
take ^= row[bit].mask;
}
unsigned long long base = A.base;
for (int i = 0; i < n; ++i) {
if ((take >> i) & 1ULL) base ^= u[i];
}
Space res = point(base);
Row depRow[61];
auto dep_add = [&](unsigned long long val, unsigned long long mask) -> unsigned long long {
for (int bit = 60; bit >= 0; --bit) {
if (!((val >> bit) & 1ULL)) continue;
if (!depRow[bit].val) {
depRow[bit] = {val, mask};
return ULLONG_MAX;
}
val ^= depRow[bit].val;
mask ^= depRow[bit].mask;
}
return mask;
};
for (int i = 0; i < n; ++i) dep_add(u[i], 1ULL << i);
for (int j = 0; j < m; ++j) {
unsigned long long dep = dep_add(v[j], 1ULL << (n + j));
if (dep == ULLONG_MAX) continue;
unsigned long long dir = 0;
for (int i = 0; i < n; ++i) {
if ((dep >> i) & 1ULL) dir ^= u[i];
}
if (dir) res.basis.add(dir);
}
return normalize(res);
}
Space unite_affine(const Space& A, const Space& B) {
if (A.empty) return B;
if (B.empty) return A;
Space res = A;
res.basis.add(A.base ^ B.base);
for (auto x : B.basis.vecs()) res.basis.add(x);
return normalize(res);
}
pair<ll, ll> part(ll l, ll r, int k, int side) {
if (l > r) return {1, 0};
ll h = 1LL << (k - 1);
ll L = side ? h : 0;
ll R = side ? 2 * h - 1 : h - 1;
l = max(l, L);
r = min(r, R);
if (l > r) return {1, 0};
if (side) l -= h, r -= h;
return {l, r};
}
struct Key {
int k;
ll a, b, c, d;
bool operator==(const Key& o) const {
return k == o.k && a == o.a && b == o.b && c == o.c && d == o.d;
}
};
struct HashKey {
size_t operator()(const Key& x) const {
unsigned long long h = x.k;
h = h * 1000003ULL + x.a;
h = h * 1000003ULL + x.b;
h = h * 1000003ULL + x.c;
h = h * 1000003ULL + x.d;
return h ^ (h >> 32);
}
};
unordered_map<Key, Space, HashKey> memo;
Space trans(int k, ll l1, ll r1, ll l2, ll r2) {
bool e1 = l1 > r1;
bool e2 = l2 > r2;
if (e1 || e2) {
if (e1 && e2) return full_space(k);
return none();
}
if (k == 0) return point(0);
Key key{k, l1, r1, l2, r2};
if (auto it = memo.find(key); it != memo.end()) return it->second;
auto a0 = part(l1, r1, k, 0);
auto a1 = part(l1, r1, k, 1);
auto b0 = part(l2, r2, k, 0);
auto b1 = part(l2, r2, k, 1);
Space keep = intersect_space(
trans(k - 1, a0.first, a0.second, b0.first, b0.second),
trans(k - 1, a1.first, a1.second, b1.first, b1.second)
);
Space flip = intersect_space(
trans(k - 1, a0.first, a0.second, b1.first, b1.second),
trans(k - 1, a1.first, a1.second, b0.first, b0.second)
);
if (!flip.empty) flip.base ^= 1ULL << (k - 1);
return memo[key] = unite_affine(keep, flip);
}
int main() {
setIO();
int t;
cin >> t;
while (t--) {
ll l, r;
cin >> l >> r;
int k = 0;
while ((1LL << k) <= r) ++k;
memo.clear();
Space ans = trans(k, l, r, l, r);
cout << ((1LL << ans.basis.rank()) - 1) << '\n';
}
}#include <bits/stdc++.h>
using namespace std;
using ll = long long;
void setIO() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
}
struct Basis {
static const int B = 61;
unsigned long long a[B]{};
bool add(unsigned long long x) {
for (int i = B - 1; i >= 0; --i) {
if (!((x >> i) & 1ULL)) continue;
if (!a[i]) {
a[i] = x;
return true;
}
x ^= a[i];
}
return false;
}
vector<unsigned long long> vecs() const {
vector<unsigned long long> v;
for (int i = 0; i < B; ++i) if (a[i]) v.push_back(a[i]);
return v;
}
int rank() const {
int res = 0;
for (int i = 0; i < B; ++i) res += (a[i] != 0);
return res;
}
};
struct Space {
bool empty = true;
unsigned long long base = 0;
Basis basis;
};
Space none() { return Space{}; }
Space point(unsigned long long x) {
Space s;
s.empty = false;
s.base = x;
return s;
}
Space full_space(int k) {
Space s = point(0);
for (int i = 0; i < k; ++i) s.basis.add(1ULL << i);
return s;
}
Space normalize(Space s) {
if (s.empty) return s;
for (int i = Basis::B - 1; i >= 0; --i) {
if (((s.base >> i) & 1ULL) && s.basis.a[i]) s.base ^= s.basis.a[i];
}
return s;
}
Space intersect_space(const Space& A, const Space& B) {
if (A.empty || B.empty) return none();
vector<unsigned long long> u = A.basis.vecs();
vector<unsigned long long> v = B.basis.vecs();
int n = (int)u.size(), m = (int)v.size();
struct Row {
unsigned long long val = 0, mask = 0;
} row[61];
auto add = [&](unsigned long long val, unsigned long long mask) -> unsigned long long {
for (int bit = 60; bit >= 0; --bit) {
if (!((val >> bit) & 1ULL)) continue;
if (!row[bit].val) {
row[bit] = {val, mask};
return ULLONG_MAX;
}
val ^= row[bit].val;
mask ^= row[bit].mask;
}
return mask;
};
for (int i = 0; i < n; ++i) add(u[i], 1ULL << i);
for (int j = 0; j < m; ++j) add(v[j], 1ULL << (n + j));
unsigned long long need = A.base ^ B.base;
unsigned long long take = 0;
for (int bit = 60; bit >= 0; --bit) {
if (!((need >> bit) & 1ULL)) continue;
if (!row[bit].val) return none();
need ^= row[bit].val;
take ^= row[bit].mask;
}
unsigned long long base = A.base;
for (int i = 0; i < n; ++i) {
if ((take >> i) & 1ULL) base ^= u[i];
}
Space res = point(base);
Row depRow[61];
auto dep_add = [&](unsigned long long val, unsigned long long mask) -> unsigned long long {
for (int bit = 60; bit >= 0; --bit) {
if (!((val >> bit) & 1ULL)) continue;
if (!depRow[bit].val) {
depRow[bit] = {val, mask};
return ULLONG_MAX;
}
val ^= depRow[bit].val;
mask ^= depRow[bit].mask;
}
return mask;
};
for (int i = 0; i < n; ++i) dep_add(u[i], 1ULL << i);
for (int j = 0; j < m; ++j) {
unsigned long long dep = dep_add(v[j], 1ULL << (n + j));
if (dep == ULLONG_MAX) continue;
unsigned long long dir = 0;
for (int i = 0; i < n; ++i) {
if ((dep >> i) & 1ULL) dir ^= u[i];
}
if (dir) res.basis.add(dir);
}
return normalize(res);
}
Space unite_affine(const Space& A, const Space& B) {
if (A.empty) return B;
if (B.empty) return A;
Space res = A;
res.basis.add(A.base ^ B.base);
for (auto x : B.basis.vecs()) res.basis.add(x);
return normalize(res);
}
pair<ll, ll> part(ll l, ll r, int k, int side) {
if (l > r) return {1, 0};
ll h = 1LL << (k - 1);
ll L = side ? h : 0;
ll R = side ? 2 * h - 1 : h - 1;
l = max(l, L);
r = min(r, R);
if (l > r) return {1, 0};
if (side) l -= h, r -= h;
return {l, r};
}
struct Key {
int k;
ll a, b, c, d;
bool operator==(const Key& o) const {
return k == o.k && a == o.a && b == o.b && c == o.c && d == o.d;
}
};
struct HashKey {
size_t operator()(const Key& x) const {
unsigned long long h = x.k;
h = h * 1000003ULL + x.a;
h = h * 1000003ULL + x.b;
h = h * 1000003ULL + x.c;
h = h * 1000003ULL + x.d;
return h ^ (h >> 32);
}
};
unordered_map<Key, Space, HashKey> memo;
Space trans(int k, ll l1, ll r1, ll l2, ll r2) {
bool e1 = l1 > r1;
bool e2 = l2 > r2;
if (e1 || e2) {
if (e1 && e2) return full_space(k);
return none();
}
if (k == 0) return point(0);
Key key{k, l1, r1, l2, r2};
if (auto it = memo.find(key); it != memo.end()) return it->second;
auto a0 = part(l1, r1, k, 0);
auto a1 = part(l1, r1, k, 1);
auto b0 = part(l2, r2, k, 0);
auto b1 = part(l2, r2, k, 1);
Space keep = intersect_space(
trans(k - 1, a0.first, a0.second, b0.first, b0.second),
trans(k - 1, a1.first, a1.second, b1.first, b1.second)
);
Space flip = intersect_space(
trans(k - 1, a0.first, a0.second, b1.first, b1.second),
trans(k - 1, a1.first, a1.second, b0.first, b0.second)
);
if (!flip.empty) flip.base ^= 1ULL << (k - 1);
return memo[key] = unite_affine(keep, flip);
}
int main() {
setIO();
int t;
cin >> t;
while (t--) {
ll l, r;
cin >> l >> r;
int k = 0;
while ((1LL << k) <= r) ++k;
memo.clear();
Space ans = trans(k, l, r, l, r);
cout << ((1LL << ans.basis.rank()) - 1) << '\n';
}
}