結果
問題 | No.2211 Frequency Table of GCD |
ユーザー |
|
提出日時 | 2023-02-10 22:30:19 |
言語 | C++17(gcc12) (gcc 12.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 51 ms / 2,000 ms |
コード長 | 3,221 bytes |
コンパイル時間 | 13,537 ms |
コンパイル使用メモリ | 310,520 KB |
最終ジャッジ日時 | 2025-02-10 13:01:13 |
ジャッジサーバーID (参考情報) |
judge5 / judge5 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 26 |
ソースコード
#pragma GCC target ("avx2")#pragma GCC optimize ("O3")#pragma GCC optimize("Ofast")#pragma GCC optimize ("unroll-loops")// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")#include <bits/stdc++.h>#include <ext/pb_ds/assoc_container.hpp>using namespace __gnu_pbds;using namespace std;#define io ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)#define endl '\n'typedef long long ll;#define mod1 (ll)998244353#define int ll#define pll pair<ll,ll>typedef long double lb;typedef tree<pair<ll, ll>,null_type,less<pair<ll, ll>>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;#define eps (lb)(1e-9)struct custom_hash {static uint64_t splitmix64(uint64_t x) {x += 0x9e3779b97f4a7c15;x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;x = (x ^ (x >> 27)) * 0x94d049bb133111eb;return x ^ (x >> 31);}size_t operator()(uint64_t x) const {static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();return splitmix64(x + FIXED_RANDOM);}};// Operator overloadstemplate<typename T1, typename T2> // cin >> pair<T1, T2>istream& operator>>(istream &istream, pair<T1, T2> &p) { return (istream >> p.first >> p.second); }template<typename T> // cin >> vector<T>istream& operator>>(istream &istream, vector<T> &v){for (auto &it : v)cin >> it;return istream;}template<typename T1, typename T2> // cout << pair<T1, T2>ostream& operator<<(ostream &ostream, const pair<T1, T2> &p) { return (ostream << p.first << " " << p.second); }template<typename T> // cout << vector<T>ostream& operator<<(ostream &ostream, const vector<T> &c) { for (auto &it : c) cout << it << " "; return ostream; }// Utility functionstemplate <typename T>void print(T &&t) { cout << t << "\n"; }template <typename T, typename... Args>void print(T &&t, Args &&... args){cout << t << " ";print(forward<Args>(args)...);}mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());ll random(ll p){ // gives random number in [0,p]return uniform_int_distribution<ll>(1<<19, p)(rng);}ll power(ll a, ll b, ll c=mod1){if(a==0){return 0;}else if(b==0||a==1){return 1;}ll p=power(a,b/2,c);p*=p; p%=c;if(b%2){p*=a;} p%=c;return p;}ll modinv(ll a, ll c=mod1){return power(a,c-2,c);}ll n,m;vector<ll>freq,dp,cnt;void solve();signed main() {io;ll t=1,n=1;// cin>>t;while (t--){solve();}return 0;}void solve(){cin>>n>>m;freq.resize(m+1);freq=vector<ll>(m+1,0);dp=cnt=freq;vector<ll>v(n);cin>>v;for(auto i:v){freq[i]++;}for(ll i(1);i<=m;++i){for(ll j(i);j<=m;j+=i){dp[i]+=freq[j];}}// cout<<freq<<endl;// cout<<dp<<endl;for(ll i(m);i>0;--i){cnt[i]=power(2,dp[i])-1;for(ll j(2*i);j<=m;j+=i){cnt[i]-=cnt[j];}cnt[i]%=mod1;cnt[i]+=mod1;cnt[i]%=mod1;}for(ll i(1);i<=m;++i){cout<<cnt[i]<<endl;}}// Do not get stuck on a single approach for long, think of multiple ways