计蒜客-MaxAnswer

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题意

求数组的一个子区间，使得 子区间之和 乘以 子区间最小值 最大，求最大值

题解

权值可能为负，分别考虑正负，正数使用但单调栈，负数求出最小连续子段和乘以该子段的最小值

代码

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#include <bits/stdc++.h>
using namespace std;
#define forl(i, l, r) for (int i = l; i <= r; i++)
#define forr(i, r, l) for (int i = r; i >= l; i--)
#define for1(i, n) for (int i = 1; i <= n; i++)
#define fro1(i, n) for (int i = 1; i <= n; i++)
#define for0(i, n) for (int i = 0; i < n; i++)
#define fro0(i, n) for (int i = 0; i < n; i++)
#define meminf(a) memset(a, inf, sizeof(a))
#define mem_1(a) memset(a, -1, sizeof(a))
#define mem0(a) memset(a, 0, sizeof(a))
#define memcp(a,b) memcpy(a,b,sizeof(b))
#define oper(type) bool operator <(const type y)const
#define mp make_pair
#define pu_b push_back
#define pu_f push_front
#define po_b pop_back
#define po_f pop_front
#define fi first
#define se second
#define whiel while
#define retrun return
typedef pair<long long, long long> pll;
typedef vector<long long> vll;
typedef pair<int, int> pii;
typedef unsigned long long ull;
typedef vector<int> vii;
typedef double db;
typedef long long ll;
typedef int itn;
int in(int &a,int &b,int &c,int &d){return scanf("%d%d%d%d",&a,&b,&c,&d);}
int in(int &a,int &b,int &c){return scanf("%d%d%d",&a,&b,&c);}
int in(int &a,int &b){return scanf("%d%d",&a,&b);}
int in(ll &a){return scanf("%lld",&a);}
int in(int &a){return scanf("%d",&a);}
int in(char *s){return scanf("%s",s);}
int in(char &s){return scanf("%c",&s);}
int in(db &a){return scanf("%lf",&a);}
void out(int a){printf("%d ",a);}
void outln(int a){printf("%d\n",a);}
void out(ll a){printf("%lld ",a);}
void outln(ll a){printf("%lld\n",a);}
const db pi = acos((db)-1);
const ll inf =0x3f3f3f3f;
const db eps = 1e-8;
const int N = 5.1e5;
const int M = 2.1e5;
const ll mod = 1e9+7;
int sign(db a) { return a < -eps ? -1 : a > eps;}
int db_cmp(db a, db b){ return sign(a-b); }


stack<pll>sta; 
int main() {
#ifdef PerpEternal
    // freopen("/Users/perpeternal/Documents/Sketch/data/in.dat", "r", stdin);
    // freopen("/Users/perpeternal/Documents/Sketch/data/out.dat", "w", stdout);
#endif
    //ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    int n;
    ll x;
    in(n);
    ll ans=0,minn=0,tot=0;
    for0(i,n){
        in(x);
        if(tot+x>0){
            tot=minn=0;
        }else{
            minn=min(minn,x);
            tot+=x;
            ans=max(ans,minn*tot);
        }
        if(x<=0){
            ll tmp=0;
            whiel(sta.size()){
                tmp+=sta.top().se;
                ans=max(ans,sta.top().fi*tmp);
                sta.pop();
            }
        }else{
            ll tmp=0;
            while(sta.size()&&x<=sta.top().fi){
                tmp+=sta.top().se;
                ans=max(ans,sta.top().fi*tmp);
                sta.pop();
            }
            sta.push(pll(x,tmp+x));
        }
    }
    ll tmp=0;
    whiel(sta.size()){
        tmp+=sta.top().se;
        ans=max(ans,sta.top().fi*tmp);
        sta.pop();
    }
    outln(ans);
    return 0;
}