[BZOJ 3745] [COCI 2015] Norma

Description

给定一个正整数序列 $a_1,a_2,\cdots,a_n$，求

\[\sum_{i=1}^n\sum_{j=i}^n(j-i+1)\min(a_i,a_{i+1},\cdots,a_j)\max(a_i,a_{i+1},\cdots,a_j)
\]

Input

第 $1$ 行，一个整数 $n$；

第 $2\dots n+1$ 行，每行一个整数表示序列 $a$。

Output

输出答案对 $10^9$ 取模后的结果。

Sample Input

Sample Output

HINT

$n \le 500000,1 \le a_i \le 10^8$

Solution

CDQ分治，从右往左枚举 $[l,mid]$ 的每个点 $i$，设 $[i,mid]$ 的最小值为 $mn$，最大值为 $mx$，同时在 $[mid+1,r]$ 中维护两个指针 $a,b$，满足 $\min[mid+1,a]\ge mn,\max[mid+1,b]\le mx$。假设 $a<b$，那么 $[mid+1,r]$ 就被分成了三块，我们分别考虑 $j$ 在每个块内的答案：

若 $mid<j\le a$，

\[\begin{eqnarray}
ans&=&mn\cdot mx\sum_{j=mid+1}^{a}(j-i+1)\\
&=&\frac{(a+mid-2i+3)(a-mid)}{2}
\end{eqnarray}
\]

若 $a< j \le b$，

\[\begin{eqnarray}
ans&=&mx\cdot\sum_{j=a+1}^b(j-i+1)\min[a+1,j]\\
&=&mx\left(\sum_{j=a+1}^bj\cdot\min[a+1,j]-(i-1)\sum_{j=a+1}^b\min[a+1,j]\right)\\
&=&mx\left(\sum_{j=a+1}^bj\cdot\min[1,j]-(i-1)\sum_{j=a+1}^b\min[1,j]\right)
\end{eqnarray}
\]

若 $b< j\le r$，

\[\begin{eqnarray}
ans&=&\sum_{j=b+1}^r(j-i+1)\min[b+1,j]\max[b+1,j]\\
&=&\sum_{j=b+1}^rj\cdot\min[b+1,j]\max[b+1,j]-(i-1)\sum_{j=b+1}^r\min[b+1,j]\max[b+1,j]\\
&=&\sum_{j=b+1}^rj\cdot\min[1,j]\max[1,j]-(i-1)\sum_{j=b+1}^r\min[1,j]\max[1,j]
\end{eqnarray}
\]

其中 $\sum_{j=a+1}^bj\cdot\min[1,j]$，$\sum_{j=a+1}^b\min[1,j]$，$\sum_{j=b+1}^rj\cdot\min[1,j]\max[1,j]$，$\sum_{j=b+1}^r\min[1,j]\max[1,j]$ 可以预处理前缀和得到。

Code

#include <cstdio>

#include <algorithm>

using std::max; using std::min;

const int N = 500005, mod = 1000000000;

int a[N], n, s1[N], s2[N], s3[N], s4[N], s5[N], s6[N], ans;

int read() {

	int x = 0; char c = getchar();

	while (c < '0' || c > '9') c = getchar();

	while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();

	return x;

}

void solve(int l, int r) {

	if (l == r) { ans = (ans + 1LL * a[l] * a[l]) % mod; return; }

	int mid = (l + r) >> 1, mn = mod, mx = 0, x;

	solve(l, mid), solve(mid + 1, r);

	for (int i = mid + 1; i <= r; ++i) {

		mn = min(mn, a[i]), mx = max(mx, a[i]);

		s1[i] = (s1[i - 1] + 1LL * i * mn) % mod;

		if ((s2[i] = s2[i - 1] + mn) >= mod) s2[i] -= mod;

		s3[i] = (s3[i - 1] + 1LL * i * mx) % mod;

		if ((s4[i] = s4[i - 1] + mx) >= mod) s4[i] -= mod;

		s5[i] = (s5[i - 1] + 1LL * i * mn % mod * mx) % mod;

		s6[i] = (s6[i - 1] + 1LL * mn * mx) % mod;

	}

	mn = mod, mx = 0;

	for (int i = mid, j = mid, k = mid; i >= l; --i) {

		mn = min(mn, a[i]), mx = max(mx, a[i]);

		while (j < r && a[j] >= mn && a[j + 1] >= mn) ++j;

		while (k < r && a[k] <= mx && a[k + 1] <= mx) ++k;

		if ((x = min(j, k))) ans = (ans + 1LL * mn * mx % mod * (((x + mid - i - i + 3LL) * (x - mid) >> 1) % mod)) % mod;

		if (j < k) ans = (ans + ((s1[k] - s1[j]) - 1LL * (i - 1) * (s2[k] - s2[j])) % mod * mx) % mod;

		if (j > k) ans = (ans + ((s3[j] - s3[k]) - 1LL * (i - 1) * (s4[j] - s4[k])) % mod * mn) % mod;

		if ((x = max(j, k)) < r) ans = (ans + s5[r] - s5[x] - 1LL * (i - 1) * (s6[r] - s6[x])) % mod;

	}

}

int main() {

	n = read();

	for (int i = 1; i <= n; ++i) a[i] = read();

	solve(1, n);

	if (ans < 0) ans += mod;

	printf("%d\n", ans);

	return 0;

}