[ABC263D] Left Right Operation

Problem Statement

You are given an integer sequence of length $N$: $A=(A_1,A_2,\ldots,A_N)$.

You will perform the following consecutive operations just once:

• Choose an integer $x$ $(0\leq x \leq N)$. If $x$ is $0$, do nothing. If $x$ is $1$ or greater, replace each of $A_1,A_2,\ldots,A_x$ with $L$.

• Choose an integer $y$ $(0\leq y \leq N)$. If $y$ is $0$, do nothing. If $y$ is $1$ or greater, replace each of $A_{N},A_{N-1},\ldots,A_{N-y+1}$ with $R$.

Print the minimum possible sum of the elements of $A$ after the operations.

Constraints

• $1 \leq N \leq 2\times 10^5$
• $-10^9 \leq L, R\leq 10^9$
• $-10^9 \leq A_i\leq 10^9$
• All values in input are integers.

---

Input

Input is given from Standard Input in the following format:

$N$ $L$ $R$
$A_1$ $A_2$ $\ldots$ $A_N$

Output

Print the answer.

---

Sample Input 1

5 4 3
5 5 0 6 3

Sample Output 1

14

If you choose $x=2$ and $y=2$, you will get $A = (4,4,0,3,3)$, for the sum of $14$, which is the minimum sum achievable.

---

Sample Input 2

4 10 10
1 2 3 4

Sample Output 2

10

If you choose $x=0$ and $y=0$, you will get $A = (1,2,3,4)$, for the sum of $10$, which is the minimum sum achievable.

---

Sample Input 3

10 -5 -3
9 -6 10 -1 2 10 -1 7 -15 5

Sample Output 3

-58

$L$, $R$, and $A_i$ may be negative.

设把区间 \([1,l-1]\) 全部变成 \(L\),把区间 \([r+1,n]\) 全部变成 \(R\)，其余不变。预处理出数列 \(a\) 的前缀和 \(s\).
此时的总价值为 $$(s_r-s_{l-1})+L(l-1)+R(n-r)$$

\[=(s_r-rR-(s_{l-1}-lL))+Rn-L \]

枚举 \(r\)，要找到最大的 \(s_{l-1}-lL\)，在枚举的同时维护最小值就可以了。

#include<bits/stdc++.h>
using namespace std;
const int N=2e5+5;
int n,l,r,a[N];
long long s[N],t[N],ans=1e18;
int main()
{
	scanf("%d%d%d",&n,&l,&r);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",a+i);
		s[i]=s[i-1]+a[i];
		ans=min(ans,1LL*i*l+1LL*(n-i)*r);
	}
	ans=min(ans,s[n]);
	for(int i=0;i<=n;i++)
		t[i]=max(t[i-1],s[i]-1LL*i*l);
	for(int i=0;i<=n;i++)
		ans=min(ans,s[i]-t[i-1]+1LL*n*r-1LL*i*r);
	printf("%lld",ans);
}