結果
問題 | No.1958 Bit Game |
ユーザー |
|
提出日時 | 2022-06-06 00:45:46 |
言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 207 ms / 2,000 ms |
コード長 | 2,996 bytes |
コンパイル時間 | 3,660 ms |
コンパイル使用メモリ | 239,960 KB |
実行使用メモリ | 5,376 KB |
最終ジャッジ日時 | 2024-09-21 04:25:19 |
合計ジャッジ時間 | 9,605 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge2 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 30 |
ソースコード
#include <bits/stdc++.h>#include <atcoder/all>typedef long long int ll;typedef long long int ull;#define MP make_pairusing namespace std;using namespace atcoder;typedef pair<ll, ll> P;const ll MOD = 998244353;// const ll MOD = 1000000007;using mint = modint998244353;// using mint = modint1000000007;const double pi = 3.1415926536;const int MAX = 2000005;long long fac[MAX], finv[MAX], inv[MAX];template<typename T> using min_priority_queue = priority_queue<T, vector<T>, greater<T>>;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;}}// 二項係数計算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 n, ll hou) {ll ret = 0;while (n) {n /= hou;ret++;}return ret;}ll ceil(ll n, ll m) {// n > 0, m > 0ll 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;}int main() {ll n, x, y;cin >> n >> x >> y;ll anum[18];ll bnum[18];for (int i = 0; i < 18; i++) {anum[i] = 0;bnum[i] = 0;}for (int i = 0; i < x; i++) {ll A;cin >> A;for (int j = 0; j < 18; j++) {anum[j] += ((A >> j) & 1);}}for (int i = 0; i < y; i++) {ll B;cin >> B;for (int j = 0; j < 18; j++) {bnum[j] += ((B >> j) & 1);}}mint ans = 0;for (int i = 0; i < 18; i++) {mint o = 0;mint z = 1;for (int j = 0; j < n; j++) {o = z * anum[i] + o * x;z = z * (x - anum[i]);z = o * (y - bnum[i]) + z * y;o = o * bnum[i];}ans += pow_mod(2, i, MOD) * o;}cout << ans.val() << endl;return 0;}