https://codeforces.com/contest/1155/problem/D

题解：DP

1、最大连续区间和=当前i前缀和-min{0-i-1前缀和}；

2、定义三个数组，分别为：

1、从没有加x；

2、已经x并且持续到当前i位置；

3、已经加x但是当前i位置不加x；

3、long long 是必要的； 

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=300000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const ll INF = 0x3f3f3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
ll ans,cnt,flag,temp,sum[N];
ll a[N];
ll dp[3][N];
char str;
struct node{};
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    //scanf("%d",&t);
    //while(t--){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++){
        scanf("%I64d",&a[i]);
    }
    ll minl= 0;
    for(int i=1;i<=n;i++){
        sum[i]=sum[i-1]+a[i];
        dp[2][i]=max(max(dp[2][i-1]+a[i],dp[1][i-1]+a[i]),0ll);
        dp[1][i]=max(max(dp[0][i-1],dp[1][i-1])+a[i]*m,0ll);
        dp[0][i]=max(sum[i]-minl,0ll);
        ans=max(max(dp[0][i],ans),max(dp[1][i],dp[2][i]));
        minl=min(minl,sum[i]);
    }
    cout<<ans<<endl;
    //}

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}