結果
問題 | No.1036 Make One With GCD 2 |
ユーザー |
|
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 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 4 |
other | AC * 41 |
ソースコード
#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)またはsegtree<S,op,e> seg(vector<S> vec)で初期化// Sは型, Sを返す関数op(S a,S b)と単位元S e()を設定する// set(int p,S x): a[p]にxを代入する// get(int p): a[p]を取得する// prod(int l,int r): [l,r)をfoldした結果を返す// all_prod(): [0,n)をfoldした結果を返す// max_right<f>(int l): bool f(S x)を渡すとl以降でf(prod(l,r))=trueとなる最大のrを返す// min_left: max_rightの逆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();}