csp20141203 集合竞价 解题报告

Solution：
对股票出价进行排序，然后按照价格递增的次序依次设定p的价格并求成交量。
1.
 //prove that the result of price(maximum--maxprice) is info[k].price:
 //If not,the nearest data that is bigger than price
 //result=maxresult & price>maxprice , conflict

2.
 //po assending as times by
 //so if maxamount is the same, use the newest

3.buy
先比较并设置maxamount值
再 买单 减

4.
sell
先 卖单 加
再比较并设置maxamount值

注意
1.出价价格的重复性
2.buy sell出价价格相同
3.数据2^64内的范围，用long long 或__int64。
而c注意用%I64d或%lld。而与购买股数有关的变量全部用long long 或__int64，不要遗漏，如min和max函数要用到long long 或__int64。
当你的程序为80分时，就是第三点没考虑。

建议：
把buy，sell合并或分开都可以，把相同价格的值分开或合并都可以，虽然后者时间效率会高一点，
但是无疑考试期间把把buy，sell合并，不把相同价格的值合并更容易写，更容易理解，也不容易写错。

 #include <iostream>
 #include <stdlib.h>
 #include <cstring>
 #include <algorithm>
 using namespace std;
 #define maxn 8000

 //The softest way to write a right code in a competition
 //Just time limite, we don't need to write the best code(in time aspect)

 struct node
 {
     double price;
     //mode: 1:sell 0:buy
     //why put sell ahead of buy?
     //sell:add  &  buy:delete    --- we need maximum
     long amount,mode;
 }info[maxn+];

 bool cmp(struct node a,struct node b)
 {
     if (a.price<b.price)
         return true;
     else if (a.price>b.price)
         return false;
     else if (a.mode>b.mode)
         return true;
     else
         return false;
 }

 __int64 min(__int64 a,__int64 b)
 {
     if (a>b)
         return b;
     else
         return a;
 }

 int main()
 {
     long g,pos,ans,i,b[maxn+],c[maxn+];
     __int64 x,y,z,maxamount;
     double a[maxn+],maxprice;
     char s[];
     bool vis[maxn+];
     //use stdlib.h faster!
     g=;
     while (scanf("%s",s)!=EOF)
     {
         g++;
         if (strcmp(s,"cancel")==)
         {
             scanf("%ld",&pos);
             vis[g]=false;
             vis[pos]=false;
         }
         else
         {
             //pay attention:%lf        double a[]
             scanf("%lf%ld",&a[g],&b[g]);
             if (strcmp(s,"sell")==)
                 c[g]=;
             else
                 c[g]=;
             vis[g]=true;
         }
     }
     ans=;
     for (i=;i<=g;i++)
         if (vis[i])
         {
             info[ans].price=a[i];
             info[ans].amount=b[i];
             info[ans].mode=c[i];
             ans++;
         }
     sort(info,info+ans,cmp);
     //prove that the result of price(maximum--maxprice) is info[k].price:
     //If not,the nearest data that is bigger than price
     //result=maxresult & price>maxprice , conflict

     //choose
     //in >= po
     //out <= po

     //po assending as times by
     //so if maxamount is the same, use the newest

     //buy
     x=;
     for (i=;i<ans;i++)
         if (info[i].mode==)
             x+=info[i].amount;
     //sell
     y=;
     maxamount=;
     for (i=;i<ans;i++)
     //when po is info[i].price
         //buy
         if (info[i].mode==)
         {
             z=min(x,y);
             if (z>=maxamount)
             {
                 maxamount=z;
                 maxprice=info[i].price;
             }
             x-=info[i].amount;
         }
         //sell
         else
         {
             y+=info[i].amount;
             z=min(x,y);
             if (z>=maxamount)
             {
                 maxamount=z;
                 maxprice=info[i].price;
             }
         }
     //数据默认maxamount>0
     printf("%.2lf %I64d",maxprice,maxamount);
     return ;
 }