結果
問題 | No.1975 Zigzag Sequence |
ユーザー |
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提出日時 | 2022-06-10 23:01:58 |
言語 | C++23 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 196 ms / 2,000 ms |
コード長 | 3,602 bytes |
コンパイル時間 | 3,207 ms |
コンパイル使用メモリ | 261,916 KB |
実行使用メモリ | 12,728 KB |
最終ジャッジ日時 | 2024-09-21 06:46:29 |
合計ジャッジ時間 | 8,081 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge4 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 33 |
ソースコード
#include <bits/stdc++.h>using namespace std;#define rep(i,n) for(int i = 0; i < (int)n; i++)#define FOR(n) for(int i = 0; i < (int)n; i++)#define repi(i,a,b) for(int i = (int)a; i < (int)b; i++)#define pb push_back#define m0(x) memset(x,0,sizeof(x))#define fill(x,y) memset(x,y,sizeof(x))#define bg begin()#define ed end()#define all(x) x.bg,x.ed//#define mp make_pair#define vi vector<int>#define vvi vector<vi>#define vll vector<ll>#define vvll vector<vll>#define vs vector<string>#define vvs vector<vs>#define vc vector<char>#define vvc vector<vc>#define pii pair<int,int>#define pllll pair<ll,ll>#define vpii vector<pair<int,int>>#define vpllll vector<pair<ll,ll>>#define vpis vector<pair<int,string>>#define vplls vector<pair<ll, string>>#define vpsi vector<pair<string, int>>#define vpsll vector<pair<string, ll>>template<typename T>void chmax(T &a, const T &b) {a = (a > b? a : b);}template<typename T>void chmin(T &a, const T &b) {a = (a < b? a : b);}using ll = long long;using ld = long double;using ull = unsigned long long;const ll INF = numeric_limits<long long>::max() / 2;const ld pi = 3.1415926535897932384626433832795028;const ll mod = 1e9 + 7;int dx[] = {-1, 0, 1, 0, -1, -1, 1, 1};int dy[] = {0, -1, 0, 1, -1, 1, -1, 1};#define int long longstruct fenwick_tree {int n, sz;vector<int> dat;function<int(int, int)> f = [](int x, int y) {return (x + y) % mod;};fenwick_tree(int n_) : dat(4*n_, 0), sz(n_) {int x = 1;while(x < n_) x *= 2;n = x;}void add(int i, int x) {i += n - 1;dat[i] += x;while(i > 0) {i = (i - 1) / 2;dat[i] = f(dat[i * 2 + 1], dat[i * 2 + 2]);}}void insert(int i) {return add(i, 1);}void erase(int i) {return add(i, -1);}void update(int i, int x) {i += n - 1;dat[i] = x;while(i > 0) {i = (i - 1) / 2;dat[i] = f(dat[i * 2 + 1], dat[i * 2 + 2]);}}int get_sum(int a, int b) {return get_sum_sub(a, b, 0, 0, n);}int get_sum_sub(int a, int b, int k, int l, int r) {if(r <= a || b <= l) {return 0;}else if(a <= l && r <= b) {return dat[k];}else {int vl = get_sum_sub(a, b, k * 2 + 1, l, (l + r) / 2);int vr = get_sum_sub(a, b, k * 2 + 2, (l + r) / 2, r);return f(vl, vr);}}int get(int i) {return dat[i + n - 1];}inline void print() {cout << "{ ";for(int i = 0; i < sz; i++) {cout << dat[i + n - 1] << " ";}cout << "}\n";}};int two[200100];void solve() {two[0] = 1;FOR(200099) two[i+1] = two[i]*2%mod;int n;cin >> n;vi a(n);FOR(n) cin >> a[i];vi val_a = a;sort(all(val_a));val_a.erase(unique(all(val_a)), val_a.end());for(auto& e : a) e = (int)(lower_bound(all(val_a), e) - val_a.begin());// for(auto e : a) cout << e << " "; cout << "\n";int ans = 0;fenwick_tree f(n+1);fenwick_tree b(n+1);FOR(n) b.add(a[i], two[n-1-i]);FOR(n) {b.add(a[i], -two[n-1-i]);ans += f.get_sum(0, a[i]) * b.get_sum(0, a[i]) % mod;ans += f.get_sum(a[i]+1, n+1) * b.get_sum(a[i]+1, n+1) % mod;ans %= mod;f.add(a[i], two[i]);}cout << ans << endl;}signed main() {cin.tie(nullptr);ios::sync_with_stdio(false);solve();return 0;}