CF 855B. Marvolo Gaunt's Ring【枚举||递推】

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Professor Dumbledore is helping Harry destroy the Horcruxes. He went to Gaunt Shack as he suspected a Horcrux to be present there. He saw Marvolo Gaunt’s Ring and identified it as a Horcrux. Although he destroyed it, he is still affected by its curse. Professor Snape is helping Dumbledore remove the curse. For this, he wants to give Dumbledore exactly x drops of the potion he made.

Value of x is calculated as maximum of p·ai + q·aj + r·ak for given p, q, r and array a1, a2, … an such that 1 ≤ i ≤ j ≤ k ≤ n. Help Snape find the value of x. Do note that the value of x may be negative.

Input
First line of input contains 4 integers n, p, q, r ( - 109 ≤ p, q, r ≤ 109, 1 ≤ n ≤ 105).

Next line of input contains n space separated integers a1, a2, … an ( - 109 ≤ ai ≤ 109).

Output
Output a single integer the maximum value of p·ai + q·aj + r·ak that can be obtained provided 1 ≤ i ≤ j ≤ k ≤ n.

Examples
input
5 1 2 3
1 2 3 4 5
output
30
input
5 1 2 -3
-1 -2 -3 -4 -5
output
12
Note
In the first sample case, we can take i = j = k = 5, thus making the answer as 1·5 + 2·5 + 3·5 = 30.

In second sample case, selecting i = j = 1 and k = 5 gives the answer 12.

一开始想到了递推式，没用数组，然后加以优化，WA在了第七组数据。
稍加研究再贴出。

AC代码：
//持续维护当前最大值

#include<iostream>#include<algorithm>#include<cmath>using namespace std;#define ll long long intconst ll _INF = -8e18;const int N = 1e5 + 10;ll dp[3][N];int a[N];int main(){    ll p, q, r, n;    cin >> n >> p >> q >> r;    for (int i = 1; i <= n; i++)    {        cin >> a[i];    }    dp[0][0] = _INF;    dp[1][0] = _INF;    dp[2][0] = _INF;    for (int i = 1; i <= n; i++)    {        dp[0][i] = max(dp[0][i - 1], p*a[i]);        dp[1][i] = max(dp[1][i - 1], dp[0][i] + q*a[i]);        dp[2][i] = max(dp[2][i - 1], dp[1][i] + r*a[i]);    }    cout << dp[2][n] << endl;    return 0;}