Codeforces 855B 简单DP

传送门：http://codeforces.com/problemset/problem/855/B

emmmmmm……n比较打，1e15直接暴力的话会T，还有要注意的一点就是选取的ai,aj,ak,下标满足i<=j<=k，orz做的时候没注意WA了好多次orz

这里要用到的是简单的DP，循环扫2遍分别从头到尾，从尾到头记录至今为止的最大最小值，（因为要求pai+qaj+rak的最大值，所以只需要p,q,r>0时*最大值，<0时*最小值就好了，问题就在于如何处理下标不交叉的问题，dp的思想就是用来处理下标和最值之间的问题）从头到尾扫一遍记录的是ai在取aj时的值，从尾到头记录的是ak在取aj时的值，然后只需要再扫一遍记录到i处pai+qaj+rak的最大值就可以了。

#include<iostream>
#include<string>
#include<string.h>
#include<cmath>
#include<algorithm>
#include<map>
using namespace std;
typedef long long ll;
ll a[100005];
int n, p[3];
ll maxx1[100005], minn1[100005];
ll maxx2[100005], minn2[100005];
int main()
{
    cin >> n;
    cin >> p[0] >> p[1] >> p[2];
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    minn1[0]=1e18;
    maxx1[0] = -1e18;
    minn2[n + 1] = 1e18;
    maxx2[n + 1] = -1e18;
    for (int i = 1; i <= n; i++)
    {
        maxx1[i] = max(maxx1[i - 1], a[i]);
        minn1[i] = min(minn1[i - 1], a[i]);
    }
    for (int i = n; i >= 1; i--)
    {
        maxx2[i] = max(maxx2[i + 1], a[i]);
        minn2[i] = min(minn2[i + 1], a[i]);
    }
    ll ans = -4e18;
    for (int i = 1; i <= n; i++)
    {
        ll tmp = 0;
        if (p[0] >= 0) tmp += p[0] * maxx1[i];
        else tmp += p[0] * minn1[i];
        if (p[2] >= 0) tmp += p[2] * maxx2[i];
        else tmp += p[2] * minn2[i];
        ans = max(ans, tmp + p[1] * a[i]);
    }
    cout << ans << endl;
    return 0;
}