結果

問題 No.1036 Make One With GCD 2
ユーザー raven7959
提出日時 2022-05-10 11:49:29
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 687 ms / 2,000 ms
コード長 5,900 bytes
コンパイル時間 2,057 ms
コンパイル使用メモリ 198,632 KB
最終ジャッジ日時 2025-01-29 05:31:20
ジャッジサーバーID
(参考情報)
judge4 / judge3
このコードへのチャレンジ
(要ログイン)
ファイルパターン 結果
sample AC * 4
other AC * 41
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>
// #pragma GCC optimize("Ofast")
// #pragma GCC optimize("unroll-loops")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,fma,abm,mmx,avx,avx2")
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rrep(i, n) for (int i = (int)(n - 1); i >= 0; i--)
#define all(x) (x).begin(), (x).end()
#define sz(x) int(x.size())
#define yn(joken) cout<<((joken) ? "Yes" : "No")<<"\n"
#define YN(joken) cout<<((joken) ? "YES" : "NO")<<"\n"
using namespace std;
using ll = long long;
using vi = vector<int>;
using vl = vector<ll>;
using vs = vector<string>;
using vc = vector<char>;
using vd = vector<double>;
using vld = vector<long double>;
using vvi = vector<vector<int>>;
using vvl = vector<vector<ll>>;
using vvs = vector<vector<string>>;
using vvc = vector<vector<char>>;
using vvd = vector<vector<double>>;
using vvld = vector<vector<long double>>;
using vvvi = vector<vector<vector<int>>>;
using vvvl = vector<vector<vector<ll>>>;
using vvvvi = vector<vector<vector<vector<int>>>>;
using vvvvl = vector<vector<vector<vector<ll>>>>;
using pii = pair<int,int>;
using pll = pair<ll,ll>;
const int INF = 1e9;
const ll LINF = 2e18;
template <class T>
bool chmax(T& a, const T& b) {
if (a < b) {
a = b;
return 1;
}
return 0;
}
template <class T>
bool chmin(T& a, const T& b) {
if (b < a) {
a = b;
return 1;
}
return 0;
}
bool ispow2(int i) { return i && (i & -i) == i; }
bool ispow2(ll i) { return i && (i & -i) == i; }
template <class T>
vector<T> make_vec(size_t a) {
return vector<T>(a);
}
template <class T, class... Ts>
auto make_vec(size_t a, Ts... ts) {
return vector<decltype(make_vec<T>(ts...))>(a, make_vec<T>(ts...));
}
template <typename T>
istream& operator>>(istream& is, vector<T>& v) {
for (int i = 0; i < int(v.size()); i++) {
is >> v[i];
}
return is;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
for (int i = 0; i < int(v.size()); i++) {
os << v[i];
if (i < int(v.size()) - 1) os << ' ';
}
return os;
}
// segtree<S,op,e> seg(int n)segtree<S,op,e> seg(vector<S> vec)
// S, Sop(S a,S b)S e()
// set(int p,S x): a[p]x
// get(int p): a[p]
// prod(int l,int r): [l,r)fold
// all_prod(): [0,n)fold
// max_right<f>(int l): bool f(S x)lf(prod(l,r))=truer
// min_left: max_right
int ceil_pow2(int n) {
int x = 0;
while ((1U << x) < (unsigned int)(n)) x++;
return x;
}
template <class S, S (*op)(S, S), S (*e)()>
struct segtree{
public:
segtree() : segtree(0) {}
segtree(int n) : segtree(vector<S>(n, e())) {}
segtree(const vector<S> &v) : _n(int(v.size())){
log = ceil_pow2(_n);
size = 1 << log;
d = vector<S>(2 * size, e());
for (int i = 0; i < _n; i++) d[size + i] = v[i];
for (int i = size - 1; i >= 1; i--) update(i);
}
void set(int p, S x){
assert(0 <= p && p < _n);
p += size;
d[p] = x;
for (int i = 1; i <= log; i++)
update(p >> i);
}
S get(int p){
assert(0 <= p && p < _n);
return d[p + size];
}
S prod(int l, int r){
assert(0 <= l && l <= r && r <= _n);
S sml = e(), smr = e();
l += size;
r += size;
while (l < r){
if (l & 1)
sml = op(sml, d[l++]);
if (r & 1)
smr = op(d[--r], smr);
l >>= 1;
r >>= 1;
}
return op(sml, smr);
}
S all_prod() { return d[1]; }
template <bool (*f)(S)>
int max_right(int l){
return max_right(l, [](S x)
{ return f(x); });
}
template <class F>
int max_right(int l, F f){
assert(0 <= l && l <= _n);
assert(f(e()));
if (l == _n) return _n;
l += size;
S sm = e();
do{
while (l % 2 == 0)
l >>= 1;
if (!f(op(sm, d[l]))){
while (l < size){
l = (2 * l);
if (f(op(sm, d[l]))){
sm = op(sm, d[l]);
l++;
}
}
return l - size;
}
sm = op(sm, d[l]);
l++;
} while ((l & -l) != l);
return _n;
}
template <bool (*f)(S)>
int min_left(int r){
return min_left(r, [](S x)
{ return f(x); });
}
template <class F>
int min_left(int r, F f){
assert(0 <= r && r <= _n);
assert(f(e()));
if (r == 0) return 0;
r += size;
S sm = e();
do{
r--;
while (r > 1 && (r % 2))
r >>= 1;
if (!f(op(d[r], sm))){
while (r < size){
r = (2 * r + 1);
if (f(op(d[r], sm))){
sm = op(d[r], sm);
r--;
}
}
return r + 1 - size;
}
sm = op(d[r], sm);
} while ((r & -r) != r);
return 0;
}
private:
int _n, size, log;
std::vector<S> d;
void update(int k) { d[k] = op(d[2 * k], d[2 * k + 1]); }
};
ll op(ll l,ll r){
return gcd(l,r);
}
ll e(){
return 0;
}
void solve(){
ll N;
cin>>N;
vl A(N);
cin>>A;
segtree<ll,op,e> seg(A);
ll ans=N*(N+1)/2;
rep(i,N){
auto f=[&](ll g)->bool{
return g!=1;
};
ll r=seg.max_right(i,f);
ans-=r-i;
}
cout<<ans<<endl;
}
int main(){
cin.tie(nullptr);
ios::sync_with_stdio(false);
solve();
}
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