2020.1.16考试总结

T1
(n le 5 imes 10^5 , a_i le 10^8)
今天就临时测了这么一道题。
考场上用单调栈水了70分，结果那30分还是因为少取模（捂脸）
正解是分治，先计算左半部分，再计算右半部分，跨区间的：在左区间用一个指针从右向左枚举，右边用一个j,k, 表示能在左区间取到最大值和最小值的最大的区间右端点，再大力讨论维护一大堆数组就OK了。
挂一下sunny_r的链接
单调站(O(n*玄学))

#include<iostream>
#include<cstdio>
#define LL long long
using namespace std;
int n, top;
LL ans;
const int N = 500010, mod = 1e9;
int da[N], xiao[N], zhan[N];
LL a[N], s[N];
inline LL read() 
{
	LL res = 0; char ch = getchar(); bool XX = false;
	for (; !isdigit(ch); ch = getchar())(ch == '-') && (XX = true);
	for (; isdigit(ch); ch = getchar())res = (res << 3) + (res << 1) + (ch ^ 48);
	return XX ? -res : res;
}
inline LL S(LL a, LL b) {return ((a + b) * (b - a + 1) / 2) % mod;}
void solve3() 
{
	top = 0;
	for (int i = 1; i <= n; ++i) 
	{
		while (top && a[i] <= a[zhan[top]])xiao[zhan[top--]] = i;
		zhan[++top] = i;
	}
	top = 0;
	for (int i = 1; i <= n; ++i) 
	{
		while (top && a[i] >= a[zhan[top]])  da[zhan[top--]] = i;
		zhan[++top] = i;
	}
	for (int i = 1; i <= n; ++i) 
	{
		if (!da[i])da[i] = n + 1;
		if (!xiao[i])xiao[i] = n + 1;
	}
	LL ma, mi;
	for (int i = 1; i <= n; ++i) 
	{
		ma = mi = i;
		while (ma <= n || mi <= n) 
		{
			if (da[ma] <= xiao[mi] && ma != n + 1)(ans += (LL)a[ma] * a[mi] % mod * S(max(ma, mi) - i + 1, da[ma] - i) % mod) %= mod, ma = da[ma];
			else (ans += (LL)a[ma] * a[mi] % mod * S(max(ma, mi) - i + 1, xiao[mi] - i) % mod) %= mod, mi = xiao[mi];
		}
	}
	cout << ans;
}
signed main() 
{
	cin >> n;
	for (int i = 1; i <= n; ++i)a[i] = read();
	solve3();
	return 0;
}

正解(O(nlogn))

#include<iostream>
#include<cstdio>
#define int long long
#define LL long long
using namespace std;
int n;
LL ans;
const int N = 500010, mod = 1e9;
int a[N], c[N][2], p[N][2], q[N][2], f[N], g[N];
inline LL read() 
{
	LL res = 0; char ch = getchar(); bool XX = false;
	for (; !isdigit(ch); ch = getchar())(ch == '-') && (XX = true);
	for (; isdigit(ch); ch = getchar())res = (res << 3) + (res << 1) + (ch ^ 48);
	return XX ? -res : res;
}
void ad(int &x, int y) { x += y; if (x >= mod) x -= mod; }
void dl(int &x, int y) { x -= y; if (x < 0) x += mod; }
int getsum(int x, int y) { return ((LL)(x + y) * (y - x + 1) >> 1) % mod; }
LL S(LL x, LL y) {return (LL)(x + y) * (y - x + 1) >> 1;}
void solve(int l, int r) 
{
	if (l == r)return (void)((ans += a[l] * a[l] % mod) %= mod);
	int mid = (l + r) >> 1;
	solve(l, mid); solve(mid + 1, r);
	c[mid][0] = c[mid][1] = a[mid];
	for (int i = mid - 1; i >= l; --i) 
	{
		c[i][0] = min(c[i + 1][0], a[i]);
		c[i][1] = max(c[i + 1][1], a[i]);
	}
	int mn = mod, mx = -mod;
	f[mid] = g[mid] = p[mid][0] = p[mid][1] = q[mid][0] = q[mid][1] = 0;
	for (int i = mid + 1; i <= r; ++i) 
	{
		mn = min(mn, a[i]); mx = max(mx, a[i]);
		f[i] = (LL)mn * mx % mod * (i - mid) % mod; (f[i] += f[i - 1]) %= mod;
		g[i] = (LL)mn * mx % mod;			   (g[i] += g[i - 1]) %= mod;
		p[i][0] = (p[i - 1][0] + mn) % mod;
		q[i][0] = (q[i - 1][0] + mx) % mod;
		p[i][1] = ((LL)mn * (i - mid) % mod + p[i - 1][1]) % mod;
		q[i][1] = ((LL)mx * (i - mid) % mod + q[i - 1][1]) % mod;
	}
	int j = mid, k = mid;
	for (int i = mid; i >= l; --i) 
	{
		while (j < r && c[i][0] < a[j + 1])++j;
		while (k < r && c[i][1] > a[k + 1])++k;
		(ans += (LL)c[i][0] * c[i][1] % mod * S(mid - i + 2, min(j, k) - i + 1) % mod) %= mod;
		(ans += ((LL)g[r] * (mid - i + 1) + f[r]) % mod) %= mod;
		(ans -= ((LL)g[max(j, k)] * (mid - i + 1) + f[max(j, k)]) % mod) %= mod;
		if (j < k) 
		{
			(ans += ((LL)p[k][0] * (mid - i + 1) % mod + p[k][1]) % mod * c[i][1]) %= mod;
			(ans -= ((LL)p[j][0] * (mid - i + 1) % mod + p[j][1]) % mod * c[i][1]) %= mod;
		} 
		else 
		{
			(ans += ((LL)q[j][0] * (mid - i + 1) % mod + q[j][1]) % mod * c[i][0]) %= mod;
			(ans -= ((LL)q[k][0] * (mid - i + 1) % mod + q[k][1]) % mod * c[i][0]) %= mod;
		}
	}
}
signed main() 
{
	cin >> n;
	for (int i = 1; i <= n; ++i)a[i] = read();
	solve(1, n);
	cout << (ans % mod + mod) % mod;
	return 0;
}