PROBLEM LINK:
Practice
Contest: Division 1
Contest: Division 2
Contest: Division 3
Contest: Division 4
Author: nskybytskyi
Testers: raysh07
Editorialist: iceknight1093
DIFFICULTY:
TBD
PREREQUISITES:
Dynamic programming, DFS
PROBLEM:
You’re given a tree on N vertices. Vertex i has the value A_i written on it.
In the easy version, A_i \leq 2.
Count the number of zig-zag sequences in the tree.
A zig-zag sequence is a sequence of vertices (v_1, v_2, \ldots, v_k) such that k \geq 2, the concatenation of the simple paths (v_i, v_{i+1}) is the full path (v_1, v_k), and the values of the vertices alternate in size.
EXPLANATION:
In the easy version, we only need to worry about A_i = 1 and A_i = 2.
This means a zig-zag sequence will alternate between 1 and 2.
We’ll solve this task using dynamic programming.
Root the tree at vertex 1.
Let \text{dp}_u denote the number of zig-zag sequences such that they start at u and go into the subtree of u.
(We’ll count length-1 sequences as zig-zag too here - it’ll make things simpler to reason about.)
Then,
- If A_u = 1, we have dp_u = 1 + \sum dp_v, where the summation is taken across all v lying in the subtree of u such that A_v = 2.
- Essentially, we’re fixing the second vertex v of the zig-zag sequence; but then we can just take any zig-zag sequence starting at v itself.
- Similarly, if A_u = 2, dp_u is obtained by summing across all v in the subtree of u with A_v = 1.
To compute these quickly, we can also store the values \text{sub}_1[u] and \text{sub}_2[u].
Here \text{sub}_1[u] stores the sum of dp_v across all v in the subtree of u such that A_v = 1.
Note that:
- \text{sub}_1[u] = \sum \text{sub}_1[c] across all children c of u.
- \text{sub}_2[u] = \sum \text{sub}_2[c] across all children c of u.
- Then, if A_u = 1 we add dp_u to \text{sub}_1[u]; otherwise we add it to \text{sub}_2[u].
- As for actually computing dp_u: note that when A_u = 1 we just have \text{dp}_u = 1 + \text{sub}_2[u], since that’s exactly the sum that we’re looking for.
Now, with the \text{dp} and \text{sub} values computed, let’s actually count zig-zag sequences.
If (v_1, v_2, \ldots, v_k) is a zig-zag sequence, there are two possibilities:
- Either v_1 or v_k is an ancestor of the other.
Sequences of this type have already been computed in either \text{dp}_{v_1} or \text{dp}_{v_k} (whichever vertex is the ancestor), by definition of the DP. - Neither v_1 nor v_2 is an ancestor of the other.
Then, there will exist a unique vertex u that is the lowest common ancestor of v_1 and v_k, and is different from both.
The last observation is useful.
Let’s fix a vertex u, and try to count sequences for which u is the lowest common ancestor.
There are, again, two possibilities:
- u is itself included in the sequence.
- Suppose A_u = 1. Then, the previous and next vertices (say x_1 and x_2) should contain 2, and lie in the subtree of u.
Further, x_1 and x_2 should come from different children of u. - If x_1 and x_2 have been fixed, the total number of sequences is then just \text{dp}_{x_1} \cdot \text{dp}_{x_2}.
- Summing this up across all valid pairs of (x_1, x_2) is now a simple exercise in combinatorics.
Note that this essentially reduces to the sum of \text{sub}_2[c_1] \cdot \text{sub}_2[c_2] across pairs (c_1, c_2) of children of u.
This can be computed quickly (i.e in linear time proportional to the number of children) in several ways - for implementation details, see the code below.
- Suppose A_u = 1. Then, the previous and next vertices (say x_1 and x_2) should contain 2, and lie in the subtree of u.
- u is not included in the sequence.
- Again, let’s fix the ‘previous’ and ‘next’ vertices of the sequence x_1 and x_2; again from different children of u.
- This time, x_1 and x_2 should have different values, since they’ll be adjacent to each other in the sequence.
- As in the previous case, this reduces to something like computing the sum of \text{sub}_1[c_1]\cdot \text{sub}_2[c_2] across pairs (c_1, c_2) of children of u, which can be computed similarly.
The overall time complexity is \mathcal{O}(N), since all the computations can be done with a single DFS.
Note that we considered length-1 subsequences to be zig-zag to make ‘gluing’ things together at the LCA simple. However, by definition they aren’t zig-zag, so subtract N from the answer to account for this.
TIME COMPLEXITY:
\mathcal{O}(N) per testcase.
CODE:
Author's code (C++)
#ifndef ATCODER_INTERNAL_MATH_HPP
#define ATCODER_INTERNAL_MATH_HPP 1
#include <utility>
#ifdef _MSC_VER
#include <intrin.h>
#endif
namespace atcoder {
namespace internal {
// @param m `1 <= m`
// @return x mod m
constexpr long long safe_mod(long long x, long long m) {
x %= m;
if (x < 0) x += m;
return x;
}
// Fast modular multiplication by barrett reduction
// Reference: https://en.wikipedia.org/wiki/Barrett_reduction
// NOTE: reconsider after Ice Lake
struct barrett {
unsigned int _m;
unsigned long long im;
// @param m `1 <= m`
explicit barrett(unsigned int m) : _m(m), im((unsigned long long)(-1) / m + 1) {}
// @return m
unsigned int umod() const { return _m; }
// @param a `0 <= a < m`
// @param b `0 <= b < m`
// @return `a * b % m`
unsigned int mul(unsigned int a, unsigned int b) const {
// [1] m = 1
// a = b = im = 0, so okay
// [2] m >= 2
// im = ceil(2^64 / m)
// -> im * m = 2^64 + r (0 <= r < m)
// let z = a*b = c*m + d (0 <= c, d < m)
// a*b * im = (c*m + d) * im = c*(im*m) + d*im = c*2^64 + c*r + d*im
// c*r + d*im < m * m + m * im < m * m + 2^64 + m <= 2^64 + m * (m + 1) < 2^64 * 2
// ((ab * im) >> 64) == c or c + 1
unsigned long long z = a;
z *= b;
#ifdef _MSC_VER
unsigned long long x;
_umul128(z, im, &x);
#else
unsigned long long x =
(unsigned long long)(((unsigned __int128)(z)*im) >> 64);
#endif
unsigned long long y = x * _m;
return (unsigned int)(z - y + (z < y ? _m : 0));
}
};
// @param n `0 <= n`
// @param m `1 <= m`
// @return `(x ** n) % m`
constexpr long long pow_mod_constexpr(long long x, long long n, int m) {
if (m == 1) return 0;
unsigned int _m = (unsigned int)(m);
unsigned long long r = 1;
unsigned long long y = safe_mod(x, m);
while (n) {
if (n & 1) r = (r * y) % _m;
y = (y * y) % _m;
n >>= 1;
}
return r;
}
// Reference:
// M. Forisek and J. Jancina,
// Fast Primality Testing for Integers That Fit into a Machine Word
// @param n `0 <= n`
constexpr bool is_prime_constexpr(int n) {
if (n <= 1) return false;
if (n == 2 || n == 7 || n == 61) return true;
if (n % 2 == 0) return false;
long long d = n - 1;
while (d % 2 == 0) d /= 2;
constexpr long long bases[3] = {2, 7, 61};
for (long long a : bases) {
long long t = d;
long long y = pow_mod_constexpr(a, t, n);
while (t != n - 1 && y != 1 && y != n - 1) {
y = y * y % n;
t <<= 1;
}
if (y != n - 1 && t % 2 == 0) {
return false;
}
}
return true;
}
template <int n> constexpr bool is_prime = is_prime_constexpr(n);
// @param b `1 <= b`
// @return pair(g, x) s.t. g = gcd(a, b), xa = g (mod b), 0 <= x < b/g
constexpr std::pair<long long, long long> inv_gcd(long long a, long long b) {
a = safe_mod(a, b);
if (a == 0) return {b, 0};
// Contracts:
// [1] s - m0 * a = 0 (mod b)
// [2] t - m1 * a = 0 (mod b)
// [3] s * |m1| + t * |m0| <= b
long long s = b, t = a;
long long m0 = 0, m1 = 1;
while (t) {
long long u = s / t;
s -= t * u;
m0 -= m1 * u; // |m1 * u| <= |m1| * s <= b
// [3]:
// (s - t * u) * |m1| + t * |m0 - m1 * u|
// <= s * |m1| - t * u * |m1| + t * (|m0| + |m1| * u)
// = s * |m1| + t * |m0| <= b
auto tmp = s;
s = t;
t = tmp;
tmp = m0;
m0 = m1;
m1 = tmp;
}
// by [3]: |m0| <= b/g
// by g != b: |m0| < b/g
if (m0 < 0) m0 += b / s;
return {s, m0};
}
// Compile time primitive root
// @param m must be prime
// @return primitive root (and minimum in now)
constexpr int primitive_root_constexpr(int m) {
if (m == 2) return 1;
if (m == 167772161) return 3;
if (m == 469762049) return 3;
if (m == 754974721) return 11;
if (m == 998244353) return 3;
int divs[20] = {};
divs[0] = 2;
int cnt = 1;
int x = (m - 1) / 2;
while (x % 2 == 0) x /= 2;
for (int i = 3; (long long)(i)*i <= x; i += 2) {
if (x % i == 0) {
divs[cnt++] = i;
while (x % i == 0) {
x /= i;
}
}
}
if (x > 1) {
divs[cnt++] = x;
}
for (int g = 2;; g++) {
bool ok = true;
for (int i = 0; i < cnt; i++) {
if (pow_mod_constexpr(g, (m - 1) / divs[i], m) == 1) {
ok = false;
break;
}
}
if (ok) return g;
}
}
template <int m> constexpr int primitive_root = primitive_root_constexpr(m);
// @param n `n < 2^32`
// @param m `1 <= m < 2^32`
// @return sum_{i=0}^{n-1} floor((ai + b) / m) (mod 2^64)
unsigned long long floor_sum_unsigned(unsigned long long n,
unsigned long long m,
unsigned long long a,
unsigned long long b) {
unsigned long long ans = 0;
while (true) {
if (a >= m) {
ans += n * (n - 1) / 2 * (a / m);
a %= m;
}
if (b >= m) {
ans += n * (b / m);
b %= m;
}
unsigned long long y_max = a * n + b;
if (y_max < m) break;
// y_max < m * (n + 1)
// floor(y_max / m) <= n
n = (unsigned long long)(y_max / m);
b = (unsigned long long)(y_max % m);
std::swap(m, a);
}
return ans;
}
} // namespace internal
} // namespace atcoder
#endif // ATCODER_INTERNAL_MATH_HPP
#ifndef ATCODER_INTERNAL_TYPE_TRAITS_HPP
#define ATCODER_INTERNAL_TYPE_TRAITS_HPP 1
#include <cassert>
#include <numeric>
#include <type_traits>
namespace atcoder {
namespace internal {
#ifndef _MSC_VER
template <class T>
using is_signed_int128 =
typename std::conditional<std::is_same<T, __int128_t>::value ||
std::is_same<T, __int128>::value,
std::true_type,
std::false_type>::type;
template <class T>
using is_unsigned_int128 =
typename std::conditional<std::is_same<T, __uint128_t>::value ||
std::is_same<T, unsigned __int128>::value,
std::true_type,
std::false_type>::type;
template <class T>
using make_unsigned_int128 =
typename std::conditional<std::is_same<T, __int128_t>::value,
__uint128_t,
unsigned __int128>;
template <class T>
using is_integral = typename std::conditional<std::is_integral<T>::value ||
is_signed_int128<T>::value ||
is_unsigned_int128<T>::value,
std::true_type,
std::false_type>::type;
template <class T>
using is_signed_int = typename std::conditional<(is_integral<T>::value &&
std::is_signed<T>::value) ||
is_signed_int128<T>::value,
std::true_type,
std::false_type>::type;
template <class T>
using is_unsigned_int =
typename std::conditional<(is_integral<T>::value &&
std::is_unsigned<T>::value) ||
is_unsigned_int128<T>::value,
std::true_type,
std::false_type>::type;
template <class T>
using to_unsigned = typename std::conditional<
is_signed_int128<T>::value,
make_unsigned_int128<T>,
typename std::conditional<std::is_signed<T>::value,
std::make_unsigned<T>,
std::common_type<T>>::type>::type;
#else
template <class T> using is_integral = typename std::is_integral<T>;
template <class T>
using is_signed_int =
typename std::conditional<is_integral<T>::value && std::is_signed<T>::value,
std::true_type,
std::false_type>::type;
template <class T>
using is_unsigned_int =
typename std::conditional<is_integral<T>::value &&
std::is_unsigned<T>::value,
std::true_type,
std::false_type>::type;
template <class T>
using to_unsigned = typename std::conditional<is_signed_int<T>::value,
std::make_unsigned<T>,
std::common_type<T>>::type;
#endif
template <class T>
using is_signed_int_t = std::enable_if_t<is_signed_int<T>::value>;
template <class T>
using is_unsigned_int_t = std::enable_if_t<is_unsigned_int<T>::value>;
template <class T> using to_unsigned_t = typename to_unsigned<T>::type;
} // namespace internal
} // namespace atcoder
#endif // ATCODER_INTERNAL_TYPE_TRAITS_HPP
#ifndef ATCODER_MODINT_HPP
#define ATCODER_MODINT_HPP 1
#include <cassert>
#include <numeric>
#include <type_traits>
#ifdef _MSC_VER
#include <intrin.h>
#endif
namespace atcoder {
namespace internal {
struct modint_base {};
struct static_modint_base : modint_base {};
template <class T> using is_modint = std::is_base_of<modint_base, T>;
template <class T> using is_modint_t = std::enable_if_t<is_modint<T>::value>;
} // namespace internal
template <int m, std::enable_if_t<(1 <= m)>* = nullptr>
struct static_modint : internal::static_modint_base {
using mint = static_modint;
public:
static constexpr int mod() { return m; }
static mint raw(int v) {
mint x;
x._v = v;
return x;
}
static_modint() : _v(0) {}
template <class T, internal::is_signed_int_t<T>* = nullptr>
static_modint(T v) {
long long x = (long long)(v % (long long)(umod()));
if (x < 0) x += umod();
_v = (unsigned int)(x);
}
template <class T, internal::is_unsigned_int_t<T>* = nullptr>
static_modint(T v) {
_v = (unsigned int)(v % umod());
}
unsigned int val() const { return _v; }
mint& operator++() {
_v++;
if (_v == umod()) _v = 0;
return *this;
}
mint& operator--() {
if (_v == 0) _v = umod();
_v--;
return *this;
}
mint operator++(int) {
mint result = *this;
++*this;
return result;
}
mint operator--(int) {
mint result = *this;
--*this;
return result;
}
mint& operator+=(const mint& rhs) {
_v += rhs._v;
if (_v >= umod()) _v -= umod();
return *this;
}
mint& operator-=(const mint& rhs) {
_v -= rhs._v;
if (_v >= umod()) _v += umod();
return *this;
}
mint& operator*=(const mint& rhs) {
unsigned long long z = _v;
z *= rhs._v;
_v = (unsigned int)(z % umod());
return *this;
}
mint& operator/=(const mint& rhs) { return *this = *this * rhs.inv(); }
mint operator+() const { return *this; }
mint operator-() const { return mint() - *this; }
mint pow(long long n) const {
assert(0 <= n);
mint x = *this, r = 1;
while (n) {
if (n & 1) r *= x;
x *= x;
n >>= 1;
}
return r;
}
mint inv() const {
if (prime) {
assert(_v);
return pow(umod() - 2);
} else {
auto eg = internal::inv_gcd(_v, m);
assert(eg.first == 1);
return eg.second;
}
}
friend mint operator+(const mint& lhs, const mint& rhs) {
return mint(lhs) += rhs;
}
friend mint operator-(const mint& lhs, const mint& rhs) {
return mint(lhs) -= rhs;
}
friend mint operator*(const mint& lhs, const mint& rhs) {
return mint(lhs) *= rhs;
}
friend mint operator/(const mint& lhs, const mint& rhs) {
return mint(lhs) /= rhs;
}
friend bool operator==(const mint& lhs, const mint& rhs) {
return lhs._v == rhs._v;
}
friend bool operator!=(const mint& lhs, const mint& rhs) {
return lhs._v != rhs._v;
}
private:
unsigned int _v;
static constexpr unsigned int umod() { return m; }
static constexpr bool prime = internal::is_prime<m>;
};
template <int id> struct dynamic_modint : internal::modint_base {
using mint = dynamic_modint;
public:
static int mod() { return (int)(bt.umod()); }
static void set_mod(int m) {
assert(1 <= m);
bt = internal::barrett(m);
}
static mint raw(int v) {
mint x;
x._v = v;
return x;
}
dynamic_modint() : _v(0) {}
template <class T, internal::is_signed_int_t<T>* = nullptr>
dynamic_modint(T v) {
long long x = (long long)(v % (long long)(mod()));
if (x < 0) x += mod();
_v = (unsigned int)(x);
}
template <class T, internal::is_unsigned_int_t<T>* = nullptr>
dynamic_modint(T v) {
_v = (unsigned int)(v % mod());
}
unsigned int val() const { return _v; }
mint& operator++() {
_v++;
if (_v == umod()) _v = 0;
return *this;
}
mint& operator--() {
if (_v == 0) _v = umod();
_v--;
return *this;
}
mint operator++(int) {
mint result = *this;
++*this;
return result;
}
mint operator--(int) {
mint result = *this;
--*this;
return result;
}
mint& operator+=(const mint& rhs) {
_v += rhs._v;
if (_v >= umod()) _v -= umod();
return *this;
}
mint& operator-=(const mint& rhs) {
_v += mod() - rhs._v;
if (_v >= umod()) _v -= umod();
return *this;
}
mint& operator*=(const mint& rhs) {
_v = bt.mul(_v, rhs._v);
return *this;
}
mint& operator/=(const mint& rhs) { return *this = *this * rhs.inv(); }
mint operator+() const { return *this; }
mint operator-() const { return mint() - *this; }
mint pow(long long n) const {
assert(0 <= n);
mint x = *this, r = 1;
while (n) {
if (n & 1) r *= x;
x *= x;
n >>= 1;
}
return r;
}
mint inv() const {
auto eg = internal::inv_gcd(_v, mod());
assert(eg.first == 1);
return eg.second;
}
friend mint operator+(const mint& lhs, const mint& rhs) {
return mint(lhs) += rhs;
}
friend mint operator-(const mint& lhs, const mint& rhs) {
return mint(lhs) -= rhs;
}
friend mint operator*(const mint& lhs, const mint& rhs) {
return mint(lhs) *= rhs;
}
friend mint operator/(const mint& lhs, const mint& rhs) {
return mint(lhs) /= rhs;
}
friend bool operator==(const mint& lhs, const mint& rhs) {
return lhs._v == rhs._v;
}
friend bool operator!=(const mint& lhs, const mint& rhs) {
return lhs._v != rhs._v;
}
private:
unsigned int _v;
static internal::barrett bt;
static unsigned int umod() { return bt.umod(); }
};
template <int id> internal::barrett dynamic_modint<id>::bt(998244353);
using modint998244353 = static_modint<998244353>;
using modint1000000007 = static_modint<1000000007>;
using modint = dynamic_modint<-1>;
namespace internal {
template <class T>
using is_static_modint = std::is_base_of<internal::static_modint_base, T>;
template <class T>
using is_static_modint_t = std::enable_if_t<is_static_modint<T>::value>;
template <class> struct is_dynamic_modint : public std::false_type {};
template <int id>
struct is_dynamic_modint<dynamic_modint<id>> : public std::true_type {};
template <class T>
using is_dynamic_modint_t = std::enable_if_t<is_dynamic_modint<T>::value>;
} // namespace internal
} // namespace atcoder
#endif // ATCODER_MODINT_HPP
#include <bits/stdc++.h>
using namespace std;
using namespace atcoder;
using mint = modint1000000007;
struct input_checker {
string buffer;
int pos;
const string all = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz";
const string number = "0123456789";
const string upper = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
const string lower = "abcdefghijklmnopqrstuvwxyz";
input_checker() {
pos = 0;
while (true) {
int c = cin.get();
if (c == -1) {
break;
}
buffer.push_back((char) c);
}
}
int nextDelimiter() {
int now = pos;
while (now < (int) buffer.size() && buffer[now] != ' ' && buffer[now] != '\n') {
now++;
}
return now;
}
string readOne() {
assert(pos < (int) buffer.size());
int nxt = nextDelimiter();
string res;
while (pos < nxt) {
res += buffer[pos];
pos++;
}
return res;
}
string readString(int minl, int maxl, const string &pattern = "") {
assert(minl <= maxl);
string res = readOne();
assert(minl <= (int) res.size());
assert((int) res.size() <= maxl);
for (int i = 0; i < (int) res.size(); i++) {
assert(pattern.empty() || pattern.find(res[i]) != string::npos);
}
return res;
}
int readInt(int minv, int maxv) {
assert(minv <= maxv);
int res = stoi(readOne());
assert(minv <= res);
assert(res <= maxv);
return res;
}
long long readLong(long long minv, long long maxv) {
assert(minv <= maxv);
long long res = stoll(readOne());
assert(minv <= res);
assert(res <= maxv);
return res;
}
auto readInts(int n, int minv, int maxv) {
assert(n >= 0);
vector<int> v(n);
for (int i = 0; i < n; ++i) {
v[i] = readInt(minv, maxv);
if (i+1 < n) readSpace();
}
return v;
}
auto readLongs(int n, long long minv, long long maxv) {
assert(n >= 0);
vector<long long> v(n);
for (int i = 0; i < n; ++i) {
v[i] = readLong(minv, maxv);
if (i+1 < n) readSpace();
}
return v;
}
void readSpace() {
assert((int) buffer.size() > pos);
assert(buffer[pos] == ' ');
pos++;
}
void readEoln() {
assert((int) buffer.size() > pos);
assert(buffer[pos] == '\n');
pos++;
}
void readEof() {
assert((int) buffer.size() == pos);
}
};
input_checker inp;
struct S {
map<int, mint> asc, desc;
};
int solve(int n, const vector<int>& a, const vector<int>& p) {
vector<vector<int>> g(n);
for (int i = 0; i < n - 1; ++i) {
g[p[i] - 1].push_back(i + 1);
}
mint total = 0;
auto dfs = [&](auto&& self, int v, int p) -> S {
vector<S> sus;
for (auto u : g[v]) {
if (u == p) {
continue;
}
sus.push_back(self(self, u, v));
}
S sv;
for (const auto& su : sus) {
for (const auto& [k1, v1] : sv.asc) {
if (k1 >= a[v]) { continue; }
for (const auto& [k2, v2] : su.asc) {
if (k2 >= a[v]) { continue; }
total += v1 * v2;
}
}
for (const auto& [k1, v1] : sv.asc) {
for (const auto& [k2, v2] : su.desc) {
if (k1 >= k2) { continue; }
total += v1 * v2;
}
}
for (const auto& [k1, v1] : sv.desc) {
for (const auto& [k2, v2] : su.asc) {
if (k1 <= k2) { continue; }
total += v1 * v2;
}
}
for (const auto& [k1, v1] : sv.desc) {
if (k1 <= a[v]) { continue; }
for (const auto& [k2, v2] : su.desc) {
if (k2 <= a[v]) { continue; }
total += v1 * v2;
}
}
for (const auto& [k2, v2] : su.asc) {
sv.asc[k2] += v2;
}
for (const auto& [k2, v2] : su.desc) {
sv.desc[k2] += v2;
}
}
for (const auto& su : sus) {
for (const auto& [k2, v2] : su.asc) {
if (k2 >= a[v]) { continue; }
total += v2;
sv.desc[a[v]] += v2;
}
for (const auto& [k2, v2] : su.desc) {
if (k2 <= a[v]) { continue; }
total += v2;
sv.asc[a[v]] += v2;
}
}
++sv.asc[a[v]];
++sv.desc[a[v]];
// cerr << "v = " << v << '\n';
// cerr << "asc:\n";
// for (const auto& [k1, v1] : sv.asc) {
// cerr << k1 << ' ' << v1.val() << '\n';
// }
// cerr << "desc:\n";
// for (const auto& [k1, v1] : sv.desc) {
// cerr << k1 << ' ' << v1.val() << '\n';
// }
// cerr << '\n';
return sv;
};
dfs(dfs, 0, -1);
return total.val();
}
int main() {
cin.tie(0)->sync_with_stdio(0);
int t = inp.readInt(1, 10'000);
inp.readEoln();
int64_t sum_n = 0;
while (t--) {
const auto n = inp.readInt(2, 100'000);
sum_n += n;
inp.readEoln();
const auto a = inp.readInts(n, 1, 2);
inp.readEoln();
vector<int> p(n - 1);
for (int i = 2; i <= n; ++i) {
if (i > 2) { inp.readSpace(); }
p[i - 2] = inp.readInt(1, i - 1);
}
inp.readEoln();
cout << solve(n, a, p) << '\n';
}
assert(sum_n <= 100'000);
inp.readEof();
}
Editorialist's code (C++)