結果

問題 No.2377 SUM AND XOR on Tree
ユーザー au7777au7777
提出日時 2023-07-09 15:41:35
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 269 ms / 4,000 ms
コード長 3,872 bytes
コンパイル時間 4,104 ms
コンパイル使用メモリ 242,320 KB
実行使用メモリ 18,560 KB
最終ジャッジ日時 2024-07-23 12:44:58
合計ジャッジ時間 11,763 ms
ジャッジサーバーID
(参考情報)
judge5 / judge4
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 2
other AC * 33
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#include <bits/stdc++.h>
#include <atcoder/all>
typedef long long int ll;
using namespace std;
typedef pair<ll, ll> P;
using namespace atcoder;
template<typename T> using min_priority_queue = priority_queue<T, vector<T>, greater<T>>;
#define USE998244353
#ifdef USE998244353
const ll MOD = 998244353;
// const double PI = 3.141592653589;
using mint = modint998244353;
#else
const ll MOD = 1000000007;
using mint = modint1000000007;
#endif
const int MAX = 20000001;
long long fac[MAX], finv[MAX], inv[MAX];
void COMinit() {
fac[0] = fac[1] = 1;
finv[0] = finv[1] = 1;
inv[1] = 1;
for (int i = 2; i < MAX; i++){
fac[i] = fac[i - 1] * i % MOD;
inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
finv[i] = finv[i - 1] * inv[i] % MOD;
}
}
long long COM(int n, int k){
if (n < k) return 0;
if (n < 0 || k < 0) return 0;
return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
ll gcd(ll x, ll y) {
if (y == 0) return x;
else if (y > x) {
return gcd (y, x);
}
else return gcd(x % y, y);
}
ll lcm(ll x, ll y) {
return x / gcd(x, y) * y;
}
ll my_sqrt(ll x) {
//
ll m = 0;
ll M = 3000000001;
while (M - m > 1) {
ll now = (M + m) / 2;
if (now * now <= x) {
m = now;
}
else {
M = now;
}
}
return m;
}
ll keta(ll num, ll arity) {
ll ret = 0;
while (num) {
num /= arity;
ret++;
}
return ret;
}
ll ceil(ll n, ll m) {
// n > 0, m > 0
ll ret = n / m;
if (n % m) ret++;
return ret;
}
ll pow_ll(ll x, ll n) {
if (n == 0) return 1;
if (n % 2) {
return pow_ll(x, n - 1) * x;
}
else {
ll tmp = pow_ll(x, n / 2);
return tmp * tmp;
}
}
vector<ll> compress(vector<ll>& v) {
// [3 5 5 6 1 1 10 1] -> [1 2 2 3 0 0 4 0]
vector<ll> u = v;
sort(u.begin(), u.end());
u.erase(unique(u.begin(),u.end()),u.end());
map<ll, ll> mp;
for (int i = 0; i < u.size(); i++) {
mp[u[i]] = i;
}
for (int i = 0; i < v.size(); i++) {
v[i] = mp[v[i]];
}
return v;
}
vector<ll> Eratosthenes( const ll N )
{
vector<bool> is_prime( N + 1 );
for( ll i = 0; i <= N; i++ )
{
is_prime[ i ] = true;
}
vector<ll> P;
for( ll i = 2; i <= N; i++ )
{
if( is_prime[ i ] )
{
for( ll j = 2 * i; j <= N; j += i )
{
is_prime[ j ] = false;
}
P.emplace_back( i );
}
}
return P;
}
pair<mint, mint> dfs(int cur, int par, int bit, vector<vector<int>>& edge, vector<int>& a) {
mint z = 0;
mint o = 0;
if ((a[cur] >> bit) & 1) {
o = 1;
}
else {
z = 1;
}
for (auto nex : edge[cur]) {
if (nex == par) continue;
auto p = dfs(nex, cur, bit, edge, a);
mint nz = 0;
mint no = 0;
// 0...
nz += p.first * z;
no += p.first * o;
// 1...
nz += p.second * o;
no += p.second * z;
// 1...
nz += p.second * z;
no += p.second * o;
z = nz;
o = no;
}
// if (bit == 0) cout << cur << ' ' << z.val() << ' ' << o.val() << '\n';
return {z, o};
}
int main() {
int n;
cin >> n;
vector<vector<int>> edge(n);
for (int i = 0; i < n - 1; i++) {
int u, v;
cin >> u >> v;
u--;
v--;
edge[u].push_back(v);
edge[v].push_back(u);
}
vector<int> a(n);
for (int i = 0; i < n; i++) cin >> a[i];
mint ans = 0;
for (int i = 0; i < 30; i++) {
auto p = dfs(0, -1, i, edge, a);
// cout << p.second.val() << '\n';
ans += p.second * (1 << i);
}
cout << ans.val() << '\n';
return 0;
}
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