广州大学第十四届ACM大学生程序设计竞赛 J 登顶

当前我们执行到第\(i\)位置的时候，我们以\(i\)结尾的字串，一共有\(i\)个。
分别是\([1,i],[2,i],[3,i]..[i,i]\)然后令单调栈\(mx,mn\)保存最大值和最小值。我们可以发现的是，以某一个点为终点后，每一个最大/小值是向前作用的。并且具有单调性，即，\([j,i]\)的最大值是\(k\)，那么\([1,j-1]\)的最大值就不会小于\(k\)。所以利用单调栈保存最大最小值和各自的位置，利用线段树更新即可。
这道题卡longlong，交了两三发发现好像不太对，计算一下发现上限\(2e19\)。（多校能这么简单就好了qwq）

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<bitset>
#include<map>
#include<regex>
#include<cstdio>
#include <iomanip>
#pragma GCC optimize(2)
#define up(i,a,b)  for(int i=a;i<b;i++)
#define dw(i,a,b)  for(int i=a;i>b;i--)
#define upd(i,a,b) for(int i=a;i<=b;i++)
#define dwd(i,a,b) for(int i=a;i>=b;i--)
#define local
typedef long long ll;
typedef unsigned long long ull;
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
ll read()
{
	char ch = getchar(); ll x = 0, f = 1;
	while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
	while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
	return x * f;
}
typedef pair<int, int> pir;
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lrt root<<1
#define rrt root<<1|1
const int N = 1e5 + 10;
int n;
int v[N];
stack<pir>mx, mn;
struct node {
	ull lazy[2];
	ull sum;
	ull mxn[2];
	node operator+(const node temp)const {
		node res;
		node now = *this;
		res.lazy[0] = res.lazy[1] = -1;
		res.mxn[0] = temp.mxn[0] + now.mxn[0];
		res.mxn[1] = temp.mxn[1] + now.mxn[1];
		res.sum = now.sum + temp.sum;
		return res;
	}
};
struct SEG {
	node tr[N << 2];
	void pushdown(int root,int l,int r)
	{
		up(i, 0, 2)
		{
			if (tr[root].lazy[i]!=-1)
			{
				ull len = (r - l + 1);
				ull lenr = len / 2ull;
				ull lenl = len - lenr;
				tr[root << 1].lazy[i] = tr[root].lazy[i];
				tr[root << 1 | 1].lazy[i] = tr[root].lazy[i];
				tr[root << 1].mxn[i] = tr[root].lazy[i] * lenl;
				tr[root << 1 | 1].mxn[i] = tr[root].lazy[i] * lenr;
				tr[root << 1].sum = tr[root].lazy[i] * tr[root << 1].mxn[i ^ 1];
				tr[root << 1 | 1].sum = tr[root].lazy[i] * tr[root << 1 | 1].mxn[i ^ 1];
				tr[root].lazy[i] = 0;
			}
		}
	}
	void pushup(int root) {
		tr[root] = tr[root << 1] + tr[root << 1 | 1];
	}
	void update(int l, int r, int root, int L, int R,int c, ull val)
	{
		if (L <= l && r <= R)
		{
			ull len = r - l + 1;
			tr[root].mxn[c] = len * val;
			tr[root].lazy[c] = val;
			tr[root].sum = val * tr[root].mxn[c ^ 1];
			return;
		}
		pushdown(root, l, r);
		int mid = (l + r) >> 1;
		if (L <= mid)update(lson, L, R, c, val);
		if (R > mid)update(rson, L, R, c, val);
		pushup(root);
	}
}TR;
int main()
{
	n = read();
	mx.push({ 0,1e9 });
	mn.push({ 0,-1 });
	upd(i, 0, n * 4)TR.tr[i].lazy[1] = TR.tr[i].lazy[0] = -1;
	ull ans = 0;
	upd(i, 1, n)
	{
		v[i] = read();
		while (mx.size() && mx.top().second < v[i])mx.pop();
		TR.update(1, n, 1, mx.top().first + 1, i, 1, (ull)v[i]);
		while (mn.size() && mn.top().second > v[i])mn.pop();
		TR.update(1, n, 1, mn.top().first + 1, i, 0, (ull)v[i]);
		mx.push({ i,v[i] });
		mn.push({ i,v[i] });
		ans += TR.tr[1].sum;
	}
	cout << ans << endl;
	return 0;
}