結果

問題 No.2075 GCD Subsequence
ユーザー fumofumofuni
提出日時 2022-09-17 09:48:32
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 397 ms / 4,000 ms
コード長 4,401 bytes
コンパイル時間 2,275 ms
コンパイル使用メモリ 209,760 KB
最終ジャッジ日時 2025-02-07 11:11:44
ジャッジサーバーID
(参考情報)
judge5 / judge3
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 * 3
other AC * 28
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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>
using namespace std;
//#pragma GCC optimize("O3")
#define rep(i,n) for(ll i=0;i<n;i++)
#define repl(i,l,r) for(ll i=(l);i<(r);i++)
#define per(i,n) for(ll i=(n)-1;i>=0;i--)
#define perl(i,r,l) for(ll i=r-1;i>=l;i--)
#define fi first
#define se second
#define ins insert
#define pqueue(x) priority_queue<x,vector<x>,greater<x>>
#define all(x) (x).begin(),(x).end()
#define CST(x) cout<<fixed<<setprecision(x)
#define rev(x) reverse(x);
using ll=long long;
using vl=vector<ll>;
using vvl=vector<vector<ll>>;
using pl=pair<ll,ll>;
using vpl=vector<pl>;
using vvpl=vector<vpl>;
const ll MOD=1000000007;
const ll MOD9=998244353;
const int inf=1e9+10;
const ll INF=4e18;
const ll dy[9]={1,0,-1,0,1,1,-1,-1,0};
const ll dx[9]={0,1,0,-1,1,-1,1,-1,0};
template <typename T> inline bool chmax(T &a, T b) {
return ((a < b) ? (a = b, true) : (false));
}
template <typename T> inline bool chmin(T &a, T b) {
return ((a > b) ? (a = b, true) : (false));
}
const int mod = MOD9;
const int max_n = 200005;
struct mint {
ll x; // typedef long long ll;
mint(ll x=0):x((x%mod+mod)%mod){}
mint operator-() const { return mint(-x);}
mint& operator+=(const mint a) {
if ((x += a.x) >= mod) x -= mod;
return *this;
}
mint& operator-=(const mint a) {
if ((x += mod-a.x) >= mod) x -= mod;
return *this;
}
mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
mint operator+(const mint a) const { return mint(*this) += a;}
mint operator-(const mint a) const { return mint(*this) -= a;}
mint operator*(const mint a) const { return mint(*this) *= a;}
mint pow(ll t) const {
if (!t) return 1;
mint a = pow(t>>1);
a *= a;
if (t&1) a *= *this;
return a;
}
bool operator==(const mint &p) const { return x == p.x; }
bool operator!=(const mint &p) const { return x != p.x; }
// for prime mod
mint inv() const { return pow(mod-2);}
mint& operator/=(const mint a) { return *this *= a.inv();}
mint operator/(const mint a) const { return mint(*this) /= a;}
};
istream& operator>>(istream& is, mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}
using vm=vector<mint>;
using vvm=vector<vm>;
struct combination {
vector<mint> fact, ifact;
combination(int n):fact(n+1),ifact(n+1) {
assert(n < mod);
fact[0] = 1;
for (int i = 1; i <= n; ++i) fact[i] = fact[i-1]*i;
ifact[n] = fact[n].inv();
for (int i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i;
}
mint operator()(int n, int k) {
if (k < 0 || k > n) return 0;
return fact[n]*ifact[k]*ifact[n-k];
}
}comb(max_n);
struct osak{
vector<long long> lpf;// least prime factor
vector<long long> prime;// prime table
osak(long long n){//linear_sieve
lpf=vector<long long>(n+1,-1);
for (int d = 2; d <= n; ++d) {
if(lpf[d]==-1){
lpf[d]=d;prime.emplace_back(d);
}
for(auto p:prime){
if(p*d>n||p>lpf[d])break;
lpf[p*d]=p;
}
}
}
map<long long,long long> factor(int n) {
map<long long,long long> factor;
while (n > 1) {
factor[lpf[n]]++;
n /= lpf[n];
}
return factor;
}
vector<long long> divisor(int N){//O(div.size())
map<long long,long long> facs=factor(N);
vector<long long> ret={1};
for(auto p:facs){
ll range=ret.size();
ll now=1;
for(int i=0;i<p.se;i++){
now*=p.fi;
for(int j=0;j<range;j++){
ret.emplace_back(ret[j]*now);
}
}
}
sort(ret.begin(),ret.end());
return ret;
}
};
int main(){
ll n;cin >> n;
vm v(1000010);
vm cov(1000010);
osak os(1000010);
rep(i,n){
ll a;cin >> a;
vl f;for(auto x:os.factor(a))f.emplace_back(x.first);
ll c=f.size();
mint plus=1;
repl(bit,1,1<<c){
ll p=1;
rep(i,c){
if(bit>>i&1)p*=f[i];
}
if(__builtin_popcount(bit)&1)plus+=cov[p];
else plus-=cov[p];
}
repl(bit,1,1<<c){
ll p=1;
rep(i,c){
if(bit>>i&1)p*=f[i];
}
cov[p]+=plus;
}
v[a]+=plus;
}
//rep(i,10)cout << cov[i] <<" ";cout << endl;
//rep(i,11)cout << v[i] <<" ";cout << endl;
mint ans=0;
for(auto p:v)ans+=p;
cout << ans << endl;
}
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