题目大意：

一个月饼店每一个小时做出月饼的花费不一样。

储存起来要钱。最多存多久。问你把全部订单做完的最少花费。

思路分析：

ans = segma( num[]*(cost[] + (i-j)*s) )

整理一下会发现式子就是  

cost[]-j*s + i*s 

对于每个订单，我们把i拿出来分析

所以也就用cost - j*s 建树。

然后在储存期间找到最小的花费即可了。

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#define lson num<<1,s,mid
#define rson num<<1|1,mid+1,e
#define maxn 2555
#define maxm 100005
#define inf 0x3f3f3f3f
using namespace std;
typedef long long LL;

int n,m;
int days[2][13]={{0,31,28,31,30,31,30,31,31,30,31,30,31},
                {0,31,29,31,30,31,30,31,31,30,31,30,31}};
string tab[] = {"","Jan","Feb","Mar","Apr","May","Jun","Jul","Aug","Sep","Oct","Nov","Dec"};
LL tre[maxm<<2];

int getmonth(string x)
{
    for(int i=1;i<=12;i++)
    if(x==tab[i])return i;
}
bool leap(int x)
{
    if(((x%4==0) && x%100!=0) || x%400==0)return true;
    return false;
}
LL gethour(int month,int day,int year,int hour)
{
    LL res=day-1;
    int is=leap(year);
    for(int i=1;i<month;i++)res+=days[is][i];
    for(int i=2000;i<year;i++)
    res+=365+leap(i);
    res*=24;
    res+=hour+1;
    return res;
}
void build(int num,int s,int e)
{
    tre[num]=inf;
    if(s==e)return;
    int mid=(s+e)>>1;
    build(lson);
    build(rson);
}
void update(int num,int s,int e,int pos,LL val)
{
    if(s==e)
    {
        tre[num]=val;
        return;
    }
    int mid=(s+e)>>1;
    if(pos<=mid)update(lson,pos,val);
    else update(rson,pos,val);
    tre[num]=min(tre[num<<1],tre[num<<1|1]);
}
LL query(int num,int s,int e,int l,int r)
{
    if(l<=s && r>=e)
    {
        return tre[num];
    }
    int mid=(s+e)>>1;
    if(r<=mid)return query(lson,l,r);
    else if(l>mid)return query(rson,l,r);
    else return min(query(lson,l,mid),query(rson,mid+1,r));
}

string tmp;
LL num[maxn];
LL cost[maxm];
LL time[maxm];
int main()
{
    while(cin>>n>>m)
    {
        if(n==0 && m==0)break;
        for(int i=1;i<=n;i++)
        {
            int d,y,h,Num;
            cin>>tmp;
            cin>>d>>y>>h>>Num;
            num[i]=Num;
            time[i]=gethour(getmonth(tmp),d,y,h);
        }
        LL S,T;
        build(1,1,m);
        cin>>T>>S;
        for(int i=1;i<=m;i++)
        {
            cin>>cost[i];
            cost[i]-=i*S;
            update(1,1,m,i,cost[i]);
        }
        LL ans=0;
        for(int i=1;i<=n;i++)
        {
            if(time[i]>m)break;
            ans+=num[i]*(query(1,1,m,max(1LL,time[i]-T+1),time[i])+time[i]*S);
        }
        cout<<ans<<endl;
    }
    return 0;
}