結果

問題 No.2211 Frequency Table of GCD
ユーザー Ujjwal RanaUjjwal Rana
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
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

#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 overloads
template<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 functions
template <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
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