結果

問題 No.1036 Make One With GCD 2
ユーザー Enjapma_kyopro
提出日時 2020-06-01 16:47:07
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 946 ms / 2,000 ms
コード長 3,794 bytes
コンパイル時間 2,184 ms
コンパイル使用メモリ 181,572 KB
実行使用メモリ 19,432 KB
最終ジャッジ日時 2024-09-16 14:23:59
合計ジャッジ時間 18,140 ms
ジャッジサーバーID
(参考情報)
judge3 / judge5
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 * 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>
#include <unistd.h>
#include <sys/time.h>
#include <stdlib.h>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
typedef pair<P, ll> T;
typedef pair<long double , ll> Ps;
const ll INF = 2e18;
const ll fact_table = 3200008;
long double Pi = 3.1415926535897932384626;
priority_queue <ll> pql;
priority_queue <P> pqp;
priority_queue <P> bag;
//big priority queue
priority_queue <ll,vector<ll>,greater<ll> > pqls;
priority_queue <P,vector<P>,greater<P> > pqps;
//small priority queue
//top pop
ll dx[8]={1,0,-1,0,1,1,-1,-1};
ll dy[8]={0,1,0,-1,1,-1,-1,1};
char dir[] = "DRUL";
//↓,→,↑,←
#define endl "\n"
#ifdef ENJAPMA
#undef endl
#endif
#define p(x) cout<<x<<endl;
#define el cout<<endl;
#define pe(x) cout<<x<<" ";
#define ps(x) cout<<fixed<<setprecision(25)<<x<<endl;
#define pu(x) cout<<(x);
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define CLEAR(a) a = decltype(a)();
#define pc(x) cout << x << ",";
#define rep(i, n) for (ll i = 0; i < (n); i ++)
// const ll mod = 998244353ll;
const ll mod = 1000000007ll;
ll mypow(ll a, ll b, ll mod){ll x=1;while(b){while(!(b&1)){(a*=a)%=mod;b>>=1;}(x*=a)%=mod;b--;}return x;}
void YES(bool cond){ if(cond){ p("YES");}else{ p("NO");} return;}
void Yes(bool cond){ if(cond){ p("Yes");}else{ p("No");} return;}
ll mygcd(ll a, ll b){
if (a < b) swap(a, b);
if (a == b || b == 0) return a;
return mygcd(b, a % b);
}
struct SegmentTree {
//SegmentTree seg( vector<ll>(l + 5, INF) );
private:
ll n;
vector<ll> node;
public:
SegmentTree(vector<ll> v) {
ll sz = v.size();
n = 1; while(n < sz) n *= 2;
node.resize(2*n-1, INF);
for(int i=0; i<sz; i++) node[i+n-1] = v[i];
for(int i=n-2; i>=0; i--) node[i] = mygcd(node[2*i+1], node[2*i+2]);
}
void update(ll x, ll val) {
//a[x]y()
x += (n - 1);
node[x] = val;
while(x > 0) {
x = (x - 1) / 2;
node[x] = mygcd(node[2*x+1], node[2*x+2]);
}
}
ll getgcd(int a, int b, int k=0, int l=0, int r=-1){
// [a, b)
if (r < 0) r = n;
if (r <= a || b <= l) return 0ll;
if (a <= l && r <= b) return node[k];
ll vl = getgcd(a, b, 2*k+1, l, (l+r)/2);
ll vr = getgcd(a, b, 2*k+2, (l+r)/2, r);
return mygcd(vl, vr);
}
};
/*
ll fact[fact_table + 5],rfact[fact_table + 5];
void c3_init(){
fact[0] = rfact[0] = 1;
for(ll i=1; i<=fact_table; i++){
fact[i] = (fact[i-1]*i) % mod;
}
rfact[fact_table] = mypow(fact[fact_table],mod - 2, mod);
for(ll i=fact_table; i>=1; i--){
rfact[i-1] = rfact[i] * i;
rfact[i-1] %= mod;
}
return;}
ll c3(ll n,ll r){
return (((fact[n] * rfact[r]) % mod ) * rfact[n-r]) % mod;}
*/
bool multicase = false;
ll n,m,num,a,b,c,d,e,h,q;
ll w, k, ans;
ll x[500005], y[500005], z[500005];
ll dp[5005][5005];
string s;
void solve(){
cin >> n;
vector<ll> v;
for(ll i=0;i<n;i++){
cin >> a;
v.pb(a);
}
SegmentTree seg(v);
ll ans = 0;
ll ok = 0;
ll right = 0;
for(ll left = 0; left < n; left ++){
while(right < n && seg.getgcd(left, right + 1) != 1){
right ++;
}
ans += n - right;
if (left == right) right ++;
}
p(ans);
return;
}
int main(){
// init();
ios::sync_with_stdio(false);
cin.tie(nullptr);
ll q, testcase = 1;
if(multicase){
cin >> q;
}else{
q = 1;
}
while(q--){
// pu("Case ");pu("#");pu(testcase);pu(": ");
solve();
testcase++;
}
// solve();
return 0;
}
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