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。
1 #include <iostream> 2 #include <stdlib.h> 3 #include <cstring> 4 #include <algorithm> 5 using namespace std; 6 #define maxn 8000 7 8 //The softest way to write a right code in a competition 9 //Just time limite, we don't need to write the best code(in time aspect) 10 11 struct node 12 { 13 double price; 14 //mode: 1:sell 0:buy 15 //why put sell ahead of buy? 16 //sell:add & buy:delete --- we need maximum 17 long amount,mode; 18 }info[maxn+1]; 19 20 bool cmp(struct node a,struct node b) 21 { 22 if (a.price<b.price) 23 return true; 24 else if (a.price>b.price) 25 return false; 26 else if (a.mode>b.mode) 27 return true; 28 else 29 return false; 30 } 31 32 __int64 min(__int64 a,__int64 b) 33 { 34 if (a>b) 35 return b; 36 else 37 return a; 38 } 39 40 int main() 41 { 42 long g,pos,ans,i,b[maxn+1],c[maxn+1]; 43 __int64 x,y,z,maxamount; 44 double a[maxn+1],maxprice; 45 char s[10]; 46 bool vis[maxn+1]; 47 //use stdlib.h faster! 48 g=0; 49 while (scanf("%s",s)!=EOF) 50 { 51 g++; 52 if (strcmp(s,"cancel")==0) 53 { 54 scanf("%ld",&pos); 55 vis[g]=false; 56 vis[pos]=false; 57 } 58 else 59 { 60 //pay attention:%lf double a[] 61 scanf("%lf%ld",&a[g],&b[g]); 62 if (strcmp(s,"sell")==0) 63 c[g]=1; 64 else 65 c[g]=0; 66 vis[g]=true; 67 } 68 } 69 ans=0; 70 for (i=1;i<=g;i++) 71 if (vis[i]) 72 { 73 info[ans].price=a[i]; 74 info[ans].amount=b[i]; 75 info[ans].mode=c[i]; 76 ans++; 77 } 78 sort(info,info+ans,cmp); 79 //prove that the result of price(maximum--maxprice) is info[k].price: 80 //If not,the nearest data that is bigger than price 81 //result=maxresult & price>maxprice , conflict 82 83 //choose 84 //in >= po 85 //out <= po 86 87 //po assending as times by 88 //so if maxamount is the same, use the newest 89 90 //buy 91 x=0; 92 for (i=0;i<ans;i++) 93 if (info[i].mode==0) 94 x+=info[i].amount; 95 //sell 96 y=0; 97 maxamount=0; 98 for (i=0;i<ans;i++) 99 //when po is info[i].price 100 //buy 101 if (info[i].mode==0) 102 { 103 z=min(x,y); 104 if (z>=maxamount) 105 { 106 maxamount=z; 107 maxprice=info[i].price; 108 } 109 x-=info[i].amount; 110 } 111 //sell 112 else 113 { 114 y+=info[i].amount; 115 z=min(x,y); 116 if (z>=maxamount) 117 { 118 maxamount=z; 119 maxprice=info[i].price; 120 } 121 } 122 //数据默认maxamount>0 123 printf("%.2lf %I64d",maxprice,maxamount); 124 return 0; 125 }
