Solution -「COCI 2014-2015 #2」「洛谷 P6406」Norma

\(\mathcal{Description}\)

  Link.

  给定 \(\{a_n\}\)，求：

\[\sum_{i=1}^n\sum_{j=i}^n(j-i+1)\min_{k=i}^j\{a_k\}\max_{k=i}^j\{a_k\}
\]

  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 \(p=\lfloor\frac{l+r}2\rfloor\)，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 \(s=\min_{u=i}^p\{a_u\}\)，\(t=\max_{u=i}^p\{a_u\}\)，\(j=\max_{u\in(p,r]}\{u|\min_{v=i}^j\{a_v\}=s\}\)，\(k=\max_{u\in(p,r]}\{u|\max_{v=i}^j\{a_v\}=t\}\)，只讨论 \(j\le k\)，发现对于将要计数的区间 \([i,x]\)，分三种情况：

• \(x\in(p,j]\)：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：

  \[pq\sum_{x\in(p,j]}(x-i+1)
  \]

• \(x\in(j,k]\)：最大值在 \([l,p]\) 中取到，而最小值会在 \((p,x]\) 中取到，则要求：

  \[q\sum_{x\in(j,k]}(x-i+1)\min_{u=p+1}^x\{a_u\}
  \]

  求和内是一个距离 \(\times\) 权值的形式，尝试预处理出值的前缀和和下标 \(\times\) 值的前缀和，就能 \(\mathcal O(1)\) 计算了，这里略过，详见代码。

• \(x\in(k,r]\)：最小值、最大值都在 \((p,x]\) 中取到，类似地求：

  \[\sum_{x\in(k,r]}(x-i+1)\min_{u=p+1}^x\{a_u\}\max_{u=p+1}^x\{a_u\}
  \]

  仍然预处理出两种前缀和，\(\mathcal O(1)\) 计算。

  综上，在分治的 \(\mathcal O(n\log n)\) 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\n", ans );
	return 0;
}