(mathcal{Description})
Link.
给定 ({a_n})，求：
[sum_{i=1}^nsum_{j=i}^n(j-i+1)min_{k=i}^j{a_k}max_{k=i}^j{a_k}
]
答案对 (10^9) 取模。
(mathcal{Solution})
挺可爱的一道题 w。
静态序列计数问题，可以考虑分治：对于 ([l,r])（(l<r)），令分割点 (p=lfloorfrac{l+r}2 floor)，对满足左端点在 ([l,p])，右端点在 ((p,r]) 的区间进行计算，然后递归两个区间继续求解。
对于本题，可以枚举区间左端点 (i=p,p-1,cdots,l)，记 (s=min_{u=i}^p{a_u})，(t=max_{u=i}^p{a_u})，(j=max_{uin(p,r]}{u|min_{v=i}^j{a_v}=s})，(k=max_{uin(p,r]}{u|max_{v=i}^j{a_v}=t})，只讨论 (jle k)，发现对于将要计数的区间 ([i,x])，分三种情况：
-
(xin(p,j])：最小值、最大值都在 ([l,p]) 中取到，那么只需要关心 (x) 的位置，故此情况对答案的贡献为：
[pqsum_{xin(p,j]}(x-i+1) ] -
(xin(j,k])：最大值在 ([l,p]) 中取到，而最小值会在 ((p,x]) 中取到，则要求：
[qsum_{xin(j,k]}(x-i+1)min_{u=p+1}^x{a_u} ]求和内是一个距离 ( imes) 权值的形式，尝试预处理出值的前缀和和下标 ( imes) 值的前缀和，就能 (mathcal O(1)) 计算了，这里略过，详见代码。
-
(xin(k,r])：最小值、最大值都在 ((p,x]) 中取到，类似地求：
[sum_{xin(k,r]}(x-i+1)min_{u=p+1}^x{a_u}max_{u=p+1}^x{a_u} ]仍然预处理出两种前缀和，(mathcal O(1)) 计算。
综上，在分治的 (mathcal O(nlog n)) 的时间内解决本题。
(mathcal{Code})
/* Clearink */
#include <cstdio>
inline int rint () {
int x = 0; char s = getchar ();
for ( ; s < '0' || '9' < s; s = getchar () );
for ( ; '0' <= s && s <= '9'; s = getchar () ) x = x * 10 + ( s ^ '0' );
return x;
}
const int MAXN = 5e5, MOD = 1e9;
int n, a[MAXN + 5], ans;
int smx[MAXN + 5], vmx[MAXN + 5];
int smn[MAXN + 5], vmn[MAXN + 5];
int sxn[MAXN + 5], vxn[MAXN + 5];
inline void chkmin ( int& a, const int b ) { b < a && ( a = b, 0 ); }
inline void chkmax ( int& a, const int b ) { a < b && ( a = b, 0 ); }
inline int mul ( const long long a, const int b ) { return a * b % MOD; }
inline int sub ( int a, const int b ) { return ( a -= b ) < 0 ? a + MOD : a; }
inline int add ( int a, const int b ) { return ( a += b ) < MOD ? a : a - MOD; }
inline void addeq ( int& a, const int b ) { ( a += b ) >= MOD && ( a -= MOD, 0 ); }
inline int sum ( const int l, const int r ) {
return r < l ? 0 : ( ( l + r ) * ( r - l + 1ll ) >> 1 ) % MOD;
}
inline void solve ( const int l, const int r ) {
if ( l == r ) return addeq ( ans, mul ( a[l], a[l] ) );
int mid = l + r >> 1;
smx[mid] = vmx[mid] = smn[mid] = vmn[mid] = sxn[mid] = vxn[mid] = 0;
for ( int i = mid + 1, mn = a[mid + 1], mx = a[mid + 1];
i <= r; ++i, chkmin ( mn, a[i] ), chkmax ( mx, a[i] ) ) {
vmx[i] = add ( vmx[i - 1], mx ),
smx[i] = add ( smx[i - 1], mul ( i - mid, mx ) );
vmn[i] = add ( vmn[i - 1], mn ),
smn[i] = add ( smn[i - 1], mul ( i - mid, mn ) );
vxn[i] = add ( vxn[i - 1], mul ( mn, mx ) ),
sxn[i] = add ( sxn[i - 1], mul ( i - mid, mul ( mn, mx ) ) );
}
for ( int i = mid, mn = a[i], mx = a[i], j = mid + 1, k = mid + 1;
i >= l; --i, chkmin ( mn, a[i] ), chkmax ( mx, a[i] ) ) {
for ( ; j <= r && mn <= a[j]; ++j );
for ( ; k <= r && a[k] <= mx; ++k );
int p = j < k ? j : k, q = j ^ k ^ p; // [mid+1,p),[p,q),[q,r].
addeq ( ans, mul ( mul ( mn, mx ), sum ( mid + 2 - i, p - i ) ) );
if ( j <= k ) { // max is constant while min will change.
addeq ( ans, mul ( mx, add (
mul ( mid - i + 1, sub ( vmn[k - 1], vmn[j - 1] ) ),
sub ( smn[k - 1], smn[j - 1] ) ) ) );
} else { // min is constant while max will change.
addeq ( ans, mul ( mn, add (
mul ( mid - i + 1, sub ( vmx[j - 1], vmx[k - 1] ) ),
sub ( smx[j - 1], smx[k - 1] ) ) ) );
}
addeq ( ans, add (
mul ( mid - i + 1, sub ( vxn[r], vxn[q - 1] ) ),
sub ( sxn[r], sxn[q - 1] ) ) );
}
solve ( l, mid ), solve ( mid + 1, r );
}
int main () {
n = rint ();
for ( int i = 1; i <= n; ++i ) a[i] = rint ();
solve ( 1, n );
printf ( "%d
", ans );
return 0;
}