Codeforces 855 B Marvolo Gaunt's Ring（dp）

题目地址
题意：给你4个数n，p，q，r，然后告诉你n个数a[]，让你求出满足最大的 p·ai + q·aj + r·ak 的i，j，k，并且i<=j<=k（重点）
思路：我们可以发现j在中间，我们只要维护一个前缀最大值和前缀最小值以及后缀最大值和后缀最小值，这样的话我们枚举这个j就好了，然后判断是正还是负，选取要最大值还是最小值，最后求出这个的最大值就好了。

#include <iostream>#include <cstring>#include <string>#include <queue>#include <vector>#include <map>#include <set>#include <cmath>#include <cstdio>#include <algorithm>#include <iomanip>#define N 100010#define M 2000010//双倍#define LL __int64#define inf 0x3f3f3f3f3f3f3f3f#define lson l,mid,ans<<1#define rson mid+1,r,ans<<1|1#define getMid (l+r)>>1#define movel ans<<1#define mover ans<<1|1using namespace std;const LL mod = 1000000007;const double eps = 0.001;LL n, a, b, c, num[N], sum;LL nummin[N], nummax[N], premin[N], premax[N];int main() {    cin.sync_with_stdio(false);    while (cin >> n >> a >> b >> c) {        nummax[0] = -1e18;        nummin[0] = 1e18;        premax[n + 1] = -1e18;        premin[n + 1] = 1e18;        for (int i = 1; i <= n; i++) {            cin >> num[i];            nummax[i] = max(nummax[i - 1], num[i]);            nummin[i] = min(nummin[i - 1], num[i]);        }        for (int i = n; i >= 1; i--) {            premax[i] = max(premax[i + 1], num[i]);            premin[i] = min(premin[i + 1], num[i]);        }        sum = -4e18;        for (int i = 1; i <= n; i++) {            LL ans = 0;            if (a >= 0) {                ans += a*nummax[i];            }            else {                ans += a*nummin[i];            }            if (c >= 0) {                ans += c*premax[i];            }            else {                ans += c*premin[i];            }            sum = max(sum, num[i] * b + ans);        }        cout << sum << endl;    }    return 0;}