PS
BOJ 1280 : 나무 심기
lickelon
2024. 9. 22. 16:38
- 문제 링크 : boj.kr/1280
- 난이도 : P4
- 태그 : 세그먼트 트리
코드
#include <bits/stdc++.h>
#define all(x) (x).begin(), (x).end()
#define INF 0x7FFFFFFF
#define MOD 1000000007
using namespace std;
using ll = long long;
using ld = long double;
using pii = pair<int,int>;
using pll = pair<ll, ll>;
template<typename T>
class segTree {
private:
ll n;
T id;
T(*merge)(T, T);
vector<T> tree;
public:
segTree(ll n, T id, T(*merge)(T, T)) {
this->n = n;
this->id = id;
this->merge = merge;
tree.resize(n*4);
}
void update(ll idx, T value) {
_update(1, 1, n, idx, value);
}
T query(ll l, ll r) {
return _query(1, 1, n, l, r);
}
private:
void _update(int node, int s, int e, int idx, T value) {
if(idx < s || e < idx) return;
if(s == e) {
tree[node] = value;
return;
}
_update(node*2, s, (s+e)/2, idx, value);
_update(node*2+1, (s+e)/2+1, e, idx, value);
tree[node] = merge(tree[node*2], tree[node*2+1]);
}
T _query(int node, int s, int e, int l, int r) {
if(l > e || r < s) return id;
if(l <= s && e <= r) return tree[node];
T lq = _query(node*2, s, (s+e)/2, l, r);
T rq = _query(node*2+1, (s+e)/2+1, e, l, r);
return merge(lq, rq);
}
};
int main()
{
ios::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
int n;
cin >> n;
segTree<pll> s(200001, pll(0, 0), [](pll a, pll b){
return pll(a.first+b.first, a.second+b.second);
});
vector<ll> arr(n);
for(auto &u : arr) {
cin >> u;
u+=1;
}
ll ans = 1;
s.update(arr[0], pll(arr[0], 1));
for(int i = 1; i < n; i++) {
auto l = s.query(1, arr[i]-1);
auto r = s.query(arr[i]+1, 200001);
ll sum = (arr[i]*l.second - l.first) % MOD;
sum += (r.first - arr[i]*r.second) % MOD;
ans *= sum % MOD;
ans %= MOD;
auto m = s.query(arr[i], arr[i]);
s.update(arr[i], pll(m.first+arr[i], m.second+1));
}
cout << ans;
return 0;
}
풀이
#12985와 같은 방식으로 풀 수 있다.
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