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  • dp--最大区间和变形-cf-1155D

    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

    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.

    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;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/shuizhidao/p/10776344.html
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