计蒜客-数组

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题意

给出长度为 $N$ 的数组 $A$, 定义

求

题解

从后往前枚举 $i$ , 用一个单调栈来存在固定 $i$ 的时候, $j\in[i,n]$ , 不同 $j$ 的 $x$ 的取值

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 long 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 = 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); }

struct da{
    int a,r,pre,i; //r 表示取 a 的 j 的区间的右端点
    da(int _a,int _r,int _p,ll _i){
        a=_a;
        r=_r;
        pre=_p;
        i=_i;
    }
    da(){}
};
stack<da>sta;
ll fun(itn a,int l,int r){
    return (r-l+1ll)*(l+r)/2%mod*a%mod;
}
int a[N];
int main() {
    itn n;
    ll ans=0;
    in(n);
    for1(i,n){
        in(a[i]);
    }
    for(int i=n;i>0;i--){
        ll r=n;
        whiel(sta.size()){
            da tmp=sta.top();
            if(tmp.a<=a[i]){
                sta.pop();
            }else{
                ll p=(tmp.a*1ll*tmp.i-tmp.a+a[i]-a[i]*1ll*i)/(tmp.a-a[i]);
                if(p>=tmp.r){
                    sta.pop();
                }else{
                    r=p;
                    break;
                }
            }
        }
        // cout<<i<<' '<<r<<"\n";
        assert(r<=n);
        // assert(r>=)
        int pre=0;
        if(sta.size()){
            da tmp=sta.top();
            // cout<<tmp.a<<' '<<tmp.i<<endl;
            pre=tmp.pre;
            // assert(r>=tmp.i);
            pre=(pre+fun(tmp.a,r-tmp.i+2,tmp.r-tmp.i+1))%mod;
        }
        // cout<<endl;
        sta.push(da(a[i],r,pre,i));
        ans=(ans+pre+fun(a[i],1,r-i+1))%mod;    
    }
    outln(ans);

    return 0;
}