結果
問題 | No.1036 Make One With GCD 2 |
ユーザー |
|
提出日時 | 2022-04-18 22:49:15 |
言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 1,286 ms / 2,000 ms |
コード長 | 2,273 bytes |
コンパイル時間 | 4,351 ms |
コンパイル使用メモリ | 230,708 KB |
実行使用メモリ | 15,444 KB |
最終ジャッジ日時 | 2024-12-30 06:14:05 |
合計ジャッジ時間 | 28,920 ms |
ジャッジサーバーID (参考情報) |
judge4 / judge2 |
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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>#include <atcoder/all>typedef long long int ll;typedef long long int ull;#define MP make_pairusing namespace std;using namespace atcoder;typedef pair<ll, ll> P;// const ll MOD = 998244353;const ll MOD = 1000000007;// using mint = modint998244353;using mint = modint1000000007;const double pi = 3.1415926536;const int MAX = 2000003;long long fac[MAX], finv[MAX], inv[MAX];template<typename T> using min_priority_queue = priority_queue<T, vector<T>, greater<T>>;void COMinit() {fac[0] = fac[1] = 1;finv[0] = finv[1] = 1;inv[1] = 1;for (int i = 2; i < MAX; i++){fac[i] = fac[i - 1] * i % MOD;inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;finv[i] = finv[i - 1] * inv[i] % MOD;}}// 二項係数計算long long COM(int n, int k){if (n < k) return 0;if (n < 0 || k < 0) return 0;return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;}ll gcd(ll x, ll y) {if (y == 0) return x;else if (y > x) {return gcd (y, x);}else return gcd(x % y, y);}ll lcm(ll x, ll y) {return x / gcd(x, y) * y;}ll my_sqrt(ll x) {ll m = 0;ll M = 3000000001;while (M - m > 1) {ll now = (M + m) / 2;if (now * now <= x) {m = now;}else {M = now;}}return m;}ll keta(ll n) {ll ret = 0;while (n) {n /= 10;ret++;}return ret;}ll ceil(ll n, ll m) {// n > 0, m > 0ll ret = n / m;if (n % m) ret++;return ret;}ll pow_ll(ll x, ll n) {if (n == 0) return 1;if (n % 2) {return pow_ll(x, n - 1) * x;}else {ll tmp = pow_ll(x, n / 2);return tmp * tmp;}}ll e() {return (ll)0;}int main() {int n;cin >> n;ll a[n + 1];segtree<ll, gcd, e> seg(n + 1);for (int i = 1; i <= n; i++) {cin >> a[i];seg.set(i, a[i]);}ll j = 2;ll ans = 0;for (ll i = 1; i <= n; i++) {while (seg.prod(i, j) != 1) {j++;if (j > n + 1) break;}if (j > n + 1) break;// gcd(i, i + 1,...,j - 1) == 1ans += n - (j - 1) + 1;}cout << ans << endl;return 0;}