codeforces Round 63-div2-D.Beautiful Array（线性动归）

原题地址

D. Beautiful Array

You are given an array aa consisting of nn integers. Beauty of array is the maximum sum of some consecutive subarray of this array (this subarray may be empty). For example, the beauty of the array [10, -5, 10, -4, 1] is 15, and the beauty of the array [-3, -5, -1] is 0.

You may choose at most one consecutive subarray of aa and multiply all values contained in this subarray by xx. You want to maximize the beauty of array after applying at most one such operation.

Input

The first line contains two integers nn and xx (1≤n≤3⋅105,−100≤x≤1001≤n≤3⋅105,−100≤x≤100) — the length of array aa and the integer xx respectively.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (−109≤ai≤109−109≤ai≤109) — the array aa.

Output

Print one integer — the maximum possible beauty of array aa after multiplying all values belonging to some consecutive subarray xx.

Examples

input

Copy

5 -2
-3 8 -2 1 -6

output

Copy

22

input

Copy

12 -3
1 3 3 7 1 3 3 7 1 3 3 7

output

Copy

42

input

Copy

5 10
-1 -2 -3 -4 -5

output

Copy

0

Note

In the first test case we need to multiply the subarray [-2, 1, -6], and the array becomes [-3, 8, 4, -2, 12] with beauty 22([-3, 8, 4, -2, 12]).

In the second test case we don't need to multiply any subarray at all.

In the third test case no matter which subarray we multiply, the beauty of array will be equal to 0.

 

对于每一个数，都有3种状态：

0：当前这个数不*x，且前面的数都没*x

1：当前这个数*x，

2：当前这个数不*x，且前面的数有*x

设dp[i][j]为前i个数在第i个数为j状态选取时能取得的最大值。

那么不难得出，

dp[i][0]依赖于dp[i-1][0],f[i][1]依赖于dp[i-1][0],dp[i-1][1],f[i][2]依赖于dp[i-1][1],dp[i-1][2]。

因为可以舍去前一部分最大值为负数的数列不要，只取一段字串，所以每一个状态还可以依赖于0。

那么动归方程为：

dp[i][0]=max(0ll,dp[i-1][0])+a[i];
dp[i][1]=max(0ll,max(dp[i-1][0],dp[i-1][1]))+a[i]*x;
dp[i][2]=max(0ll,max(dp[i-1][1],dp[i-1][2]))+a[i];

比较每个i的每一个状态，取其最大值，就时最终答案。

#include <iostream>
#include <stdio.h>
using namespace std;
int n,x;
long long a[300010];
long long dp[300010][5];

int main()
{
    std::ios::sync_with_stdio(false);
    cin>>n>>x;
    for(int i=1;i<=n;i++) cin>>a[i];
    long long ans=0;
    for(int i=1;i<=n;i++)
    {
        dp[i][0]=max(0ll,dp[i-1][0])+a[i];
        dp[i][1]=max(0ll,max(dp[i-1][0],dp[i-1][1]))+a[i]*x;
        dp[i][2]=max(0ll,max(dp[i-1][1],dp[i-1][2]))+a[i];
        ans=max(ans,max(dp[i][0],max(dp[i][1],dp[i][2])));
    }
    cout<<ans<<endl;
    return 0;
}

 

转载于:https://www.cnblogs.com/BakaCirno/p/10758491.html