| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 1138031 | Ekber_Ekber | 말 (IOI15_horses) | C++20 | 0 ms | 0 KiB |
#include "horses.h"
#include <bits/stdc++.h>
#define int long long
#define itn int
#define endl "\n"
#define ff first
#define ss second
#define pb push_back
#define ppb pop_back
#define ins insert
#define lb lower_bound
#define ub upper_bound
#define bs binary_search
#define count1 __builtin_popcount
#define all(v) v.begin(), v.end()
using namespace std;
struct point{
int ans, sum;
};
constexpr int MAX = 5e+5 + 3, INF = 2e+9, MOD = 1e+9 + 7, K = log2(MAX);
vector <int> a, b;
vector <point> t(4 * MAX);
void build(int v, int tl, int tr) {
if (tl == tr) {
t[v].ans = (a[tl] * b[tl]) % MOD;
t[v].sum = a[tl];
return;
}
int tm = (tl + tr) / 2;
build(v*2, tl, tm);
build(v*2+1, tm+1, tr);
t[v].sum = (t[v*2].sum * t[v*2+1].sum) % MOD;
t[v].ans = max(t[v*2].ans, (t[v*2].sum * t[v*2+1].ans) % MOD);
}
point find(int v, int tl, int tr, int l, int r) {
if (l > r) return {INF, INF};
if (tl == l && tr == r) return t[v];
int tm = (tl + tr) / 2;
point res1 = find(v*2, tl, tm, l, min(r, tm));
point res2 = find(v*2+1, tm+1, tr, max(l, tm+1), r);
point res;
res.sum = (res1.sum * res2.sum) % MOD;
res.ans = max(res1.ans, (res1.sum * res2.ans) % MOD);
return res;
}
void updatex(int v, int tl, int tr, int i, int x) {
if (tl == tr) {
t[v].ans = (x * b[tl]) % MOD;
t[v].sum = x;
a[tl] = x;
return;
}
int tm = (tl + tr) / 2;
if (i <= tm) {
updatex(v*2, tl, tm, i, x);
}
else {
updatex(v*2+1, tm+1, tr, i, x);
}
t[v].sum = (t[v*2].sum * t[v*2+1].sum) % MOD;
t[v].ans = max(t[v*2].ans, (t[v*2].sum * t[v*2+1].ans) % MOD);
}
void updatey(int v, int tl, int tr, int i, int y) {
if (tl == tr) {
t[v].ans = (a[tl] * y) % MOD;
b[tl] = y;
return;
}
int tm = (tl + tr) / 2;
if (i <= tm) {
updatey(v*2, tl, tm, i, y);
}
else {
updatey(v*2+1, tm+1, tr, i, y);
}
t[v].sum = (t[v*2].sum * t[v*2+1].sum) % MOD;
t[v].ans = max(t[v*2].ans, (t[v*2].sum * t[v*2+1].ans) % MOD);
}
int init(int N, int X[], int Y[]) {
for (int i=0; i < N; i++) {
a.pb(X[i]);
b.pb(Y[i]);
}
build(1, 0, N-1);
return 0;
}
int updateX(int pos, int val) {
return 0;
}
int updateY(int pos, int val) {
return 0;
}