# BZOJ3745 / SP22343 NORMA2 - Norma 分治，CDQ分治

• $p∈[mid + 1, p1 - 1]$，其中$p1$是第一次出现比$maxw$大或者比$minw$小的数的位置。

• $p∈[p1, p2 - 1]$，其中$p2$是第二次出现比$maxw$大或者比$minw$小的数的位置。

• $p∈[p2, r]$$r$是当前枚举区间的右端点。

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
const int N = 500000 + 5;
const LL Mod = 1000000000;
#define min(x,y) (x < y ? x : y)
#define max(x,y) (x > y ? x : y)
#define mul(x,y) ((1ll * (x % Mod) * (y % Mod)) % Mod)
#define add(x,y) ((0ll + (x % Mod) + (y % Mod)) % Mod)

int n, arr[N]; LL ans;

int _maxw[N], _minw[N];

LL mn1[N], mn2[N], mx1[N], mx2[N], mnmx1[N], mnmx2[N];

int get_maxp (int w, int l, int r) {
if (_maxw[r] <= w) return r + 1;
while (l < r) {
int mid = (l + r) >> 1;
if (_maxw[mid] > w) {
r = mid;
} else {
l = mid + 1;
}
}
return r;
}

int get_minp (int w, int l, int r) {
if (_minw[r] >= w) return r + 1;
while (l < r) {
int mid = (l + r) >> 1;
if (_minw[mid] < w) {
r = mid;
} else {
l = mid + 1;
}
}
return r;
}

void cdq (int l, int r) {
if (l == r) {
ans = add (ans, mul (arr[l], arr[r]));
return;
}
int mid = (l + r) >> 1;
cdq (l, mid + 0);
cdq (mid + 1, r);
int maxw = arr[mid], minw = arr[mid];
mx1[mid - 1] = mx2[mid - 1] = 0;
mn1[mid - 1] = mn2[mid - 1] = 0;
mnmx1[mid - 1] = mnmx2[mid - 1] = 0;
_maxw[mid] = _minw[mid] = arr[mid];
for (int i = mid + 1; i <= r; ++i) {
_maxw[i] = max (_maxw[i - 1], arr[i]);
_minw[i] = min (_minw[i - 1], arr[i]);
}
for (int p = mid; p <= r; ++p) {
mx1[p] = add (mx1[p - 1], mul (_maxw[p], (p + 1)));
mx2[p] = add (mx2[p - 1], _maxw[p]);
mn1[p] = add (mn1[p - 1], mul (_minw[p], (p + 1)));
mn2[p] = add (mn2[p - 1], _minw[p]);
mnmx1[p] = add (mnmx1[p - 1], mul (_maxw[p], mul (_minw[p], (p + 1))));
mnmx2[p] = add (mnmx2[p - 1], mul (_maxw[p], _minw[p]));
}
for (int i = mid; i >= l; --i) {
maxw = max (maxw, arr[i]);
minw = min (minw, arr[i]);
int p1 = get_maxp (maxw, mid + 1, r); // [mid + 1, r]内第一个比maxw大的地方
int p2 = get_minp (minw, mid + 1, r); // [mid + 1, r]内第一个比minw小的地方
if (p1 > p2) swap (p1, p2); // 不关注大小，主要看划分
//		cout << p1 << " " << p2 << endl;
ans = add (ans, mul (1ll * (p1 - mid - 1) * (p1 + mid - i * 2 + 2) / 2, mul (minw, maxw))); // Part 1
if (arr[p1] > maxw) {
ans = add (ans, mul (minw, add (add (mx1[p2 - 1], -mx1[p1 - 1]), -mul (i, add (mx2[p2 - 1], -mx2[p1 - 1])))));
} else {
ans = add (ans, mul (maxw, add (add (mn1[p2 - 1], -mn1[p1 - 1]), -mul (i, add (mn2[p2 - 1], -mn2[p1 - 1])))));
}
if (p2 <= r) {
}
}
}

signed main () {
//	freopen ("data.in", "r", stdin);
cin >> n;
for (int i = 1; i <= n; ++i) {
cin >> arr[i];
}
cdq (1, n);
//	cout << ans << endl;
cout << (((ans % Mod) + Mod) % Mod) << endl;;
}

posted @ 2019-05-22 17:08  maomao9173  阅读(142)  评论(0编辑  收藏  举报