dp--最大区间和变形-cf-1155D

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原文链接：http://www.cnblogs.com/shuizhidao/p/10776344.html

本文详细解析了Codeforces题目1155D.Beautiful Array的解题思路，该题要求在给定整数数组中找到最大连续子数组和，允许乘以特定值优化数组美度。文章提供了状态转移方程及C++代码实现。

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dp--最大区间和变形-cf-1155D

D. Beautiful Array

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 xxrespectively.

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

input

Copy

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

output

Copy

input

Copy

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

output

Copy

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.

状态转移方程如下

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <sstream>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <queue>
#include <iomanip>
#include <stack>
#include <climits>
#include <fstream>

using namespace std;

typedef long long LL;
const int INF = 0x3f3f3f3f;
const int N = 300005;
const int MOD = 1e9 + 7;

#define F(i, l, r) for(int i = l;i <= (r);++i)
#define RF(i, l, r) for(int i = l;i >= (r);--i)

int main()
{
    LL n, m, t, ans = 0, p = 0, sum1 = 0, sum2 = 0;
    cin >> n >> m;
    F(i, 1, n)
    {
        cin >> t;
        sum1 = max(sum1 + t, 0ll);//d[i][0] = max(0ll,d[i-1][0]+a[i]);
        sum2 = max(sum2 + m * t, sum1);//d[i][1] = max( 0ll, max(d[i-1][0],d[i-1][1])+a[i]*x);
        p = max(p + t, sum2);// d[i][2] = max( 0ll, max(d[i-1][0],  max(d[i-1][1],d[i-1][2]) )+ a[i]);
        ans = max(ans, p);
    }
    cout << ans << endl;
    return 0;
}

转载于:https://www.cnblogs.com/shuizhidao/p/10776344.html