牛客 545A 小A与最大子段和 & CF 660F Bear and Bowling 4

大意: 给定序列$a$, 求选择一个子区间$[l,r]$, 使得$\sum\limits_{i=l}^r(i-l+1)a_i$最大.

$n\le2e5, |a_i|\le 1e7$.

 

记$s[i]=\sum a[i], m[i]=\sum ia[i]$, $dp[i]$为以$i$为右端点的答案, 有

$\begin{align} \notag dp[i] & =\max\limits_{0\le j<i}\{m[i]-m[j]-j(s[i]-s[j])\} \\ & =  m[i]-\min\limits_{0\le j<i}\{m[j]-js[j]+js[i]\} \notag\end{align}$ 

然后斜率优化.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, P2 = 998244353, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head




const int N = 1e6+10;
int n, a[N], q[N];
ll s[N], m[N], dp[N], g[N];
double slope(int i, int j) {
	return ((double)g[i]-g[j])/(i-j);
}

int main() {
	scanf("%d", &n);
	REP(i,1,n) { 
		scanf("%d", a+i);
		s[i] = s[i-1]+a[i];
		m[i] = m[i-1]+(ll)i*a[i];
		g[i] = (ll)i*s[i]-m[i];
	}
	q[++*q] = 0;
	ll ans = -1e18;
	REP(i,1,n) {
		int opt=1,l=2,r=*q;
		while (l<=r) {
			if (slope(q[mid],q[mid-1])>=s[i]) opt=mid,l=mid+1;
			else r=mid-1;
		}
		ans = max(ans, m[i]-m[q[opt]]-(ll)q[opt]*(s[i]-s[q[opt]]));
		while (*q>1&&slope(i,q[*q])>slope(q[*q],q[*q-1])) --*q;
		q[++*q] = i;
	}
	printf("%lld\n", ans);
}