首页 > 代码库 > CTU Open 2008（未完工）

CTU Open 2008（未完工）

2024-07-04 17:54:46   222人阅读

链接：CTU Open 2008

【2014/05/15】今晚做了一下CTU 2008的这套题，最后的rank是3道题。基本是水题啊，我们做的是4个小时，如果完整做，我想应该还会出掉D题，主要是D题很繁琐，weikd写起都说烦。

 

 

A - Alea iacta est

B - On-Line Banking

【题意】纯模拟题，模拟银行存钱，取钱，转钱的操作。注意下细节~：

  1 #include <cstdio>
  2 #include <cstring>
  3 #include <algorithm>
  4 #include <map>
  5 #include <string>
  6 #define EPS 1e-8
  7 using namespace std;
  8 
  9 char str[1005];
 10 char cmd[1005];
 11 char name[1005],name1[1005];
 12 map<string,double> ac;
 13 
 14 void create(string id)
 15 {
 16     printf("create: ");
 17     if (ac.find(id)!=ac.end())
 18     {
 19         printf("already exists\n");
 20     }
 21     else
 22     {
 23         ac[id] = 0;
 24         printf("ok\n");
 25     }
 26 }
 27 
 28 void deposit(string id, double x)
 29 {
 30     printf("deposit %.2f: ",x);
 31     if (ac.find(id)==ac.end())
 32     {
 33         printf("no such account\n");
 34     }
 35     else
 36     {
 37         ac[id] = ac[id]+x;
 38         printf("ok\n");
 39     }
 40 }
 41 
 42 void withdraw(string id,double x)
 43 {
 44     printf("withdraw %.2f: ",x);
 45     if (ac.find(id)==ac.end())
 46     {
 47         printf("no such account\n");
 48     }
 49     else
 50     {
 51         if (ac[id]+EPS<x)
 52         {
 53             printf("insufficient funds\n");
 54         }
 55         else
 56         {
 57             ac[id]=ac[id]-x;
 58             printf("ok\n");
 59         }
 60     }
 61 }
 62 
 63 void transfer(string ida,string idb,double x)
 64 {
 65     printf("transfer %.2f: ",x);
 66     if (ac.find(ida)==ac.end()||ac.find(idb)==ac.end())
 67     {
 68         printf("no such account\n");
 69     }
 70     else
 71     {
 72         if (ida==idb)
 73         {
 74             printf("same account\n");
 75         }
 76         else
 77         {
 78             if (ac[ida]+EPS<x)
 79             {
 80                 printf("insufficient funds\n");
 81             }
 82             else
 83             {
 84                 ac[ida]=ac[ida]-x;
 85                 ac[idb]=ac[idb]+x;
 86                 char ia = *ida.rbegin();
 87                 char ib = *idb.rbegin();
 88                 if (ia==ib)
 89                 {
 90                     printf("ok\n");
 91                 }
 92                 else
 93                 {
 94                     printf("interbank\n");
 95                 }
 96             }
 97         }
 98     }
 99 }
100 
101 int main()
102 {
103     #ifdef HotWhite
104     freopen("in.txt", "r", stdin);
105     #endif
106     int n;
107     double mn;
108     while(scanf("%d",&n))
109     {
110         if (!n) break;
111         ac.clear();
112         for (int i = 0; i < n; i++)
113         {
114             scanf("%s %lf",str,&mn);
115             ac[str]=mn;
116         }
117         string s1,s2;
118         while(scanf("%s",cmd))
119         {
120             if (strcmp(cmd,"end")==0) break;
121             if (strcmp(cmd,"create")==0)
122             {
123                 scanf("%s",name);
124                 s1=name;
125                 create(s1);
126             }
127             if (strcmp(cmd,"deposit")==0)
128             {
129                 scanf("%s%lf",name,&mn);
130                 s1=name;
131                 deposit(s1,mn);
132             }
133             if (strcmp(cmd,"withdraw")==0)
134             {
135                 scanf("%s%lf",name,&mn);
136                 s1=name;
137                 withdraw(s1,mn);
138             }
139             if (strcmp(cmd,"transfer")==0)
140             {
141                 scanf("%s%s%lf",name,name1,&mn);
142                 s1=name;
143                 s2=name1;
144                 transfer(s1,s2,mn);
145             }
146         }
147         printf("end\n\n");
148     }
149     printf("goodbye\n");
150     return 0;
151 }
View Code

C - International Collegiate Programming Contest

D - Careful Declaration

E - Stock Exchange

【题意】每一组case给你一些买家的姓名和最高出价，还有卖家的姓名和最低售价。然后计算出所有成交信息。
【思路】水题，直接暴力模拟，比较价钱即可。代码如下：

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 using namespace std;
 5 const int MAX = 1010;
 6 
 7 struct BID
 8 {
 9     char name[25], type[5];
10     double price;
11 };
12 BID bid[MAX], buy[MAX], sell[MAX];
13 
14 int main() {
15 
16 #ifdef HotWhite
17     freopen("in.txt", "r", stdin);
18 #endif
19 
20     int N, flag;
21     char issuer[12];
22 
23     while(scanf("%d%s", &N, issuer) != EOF && N)
24     {
25         int cntb = 0, cnts = 0;
26         for(int i=0; i<N; ++i)
27         {
28             scanf("%s%s%lf", bid[i].name, bid[i].type, &bid[i].price);
29             if(strcmp(bid[i].type, "buy") == 0)
30             {
31                 buy[cntb++] = bid[i];
32             }
33             if(strcmp(bid[i].type, "sell") == 0)
34             {
35                 sell[cnts++] = bid[i];
36             }
37         }
38         printf("%s\n", issuer);
39         for(int i=0; i<N; ++i)
40         {
41             flag = 0;
42             printf("%s:", bid[i].name);
43             if(strcmp(bid[i].type, "buy") == 0)
44             {
45                 for(int j=0; j<cnts; ++j)
46                 {
47                     if(sell[j].price <= bid[i].price)
48                     {
49                         printf(" %s", sell[j].name);
50                         flag = 1;
51                     }
52                 }
53             }
54             if(strcmp(bid[i].type, "sell") == 0)
55             {
56                 for(int j=0; j<cntb; ++j)
57                 {
58                     if(buy[j].price >= bid[i].price)
59                     {
60                         printf(" %s", buy[j].name);
61                         flag = 1;
62                     }
63                 }
64             }
65             if(!flag)
66                 printf(" NO-ONE");
67             printf("\n");
68         }
69     }
70     return 0;
71 }
View Code

 

F - Wooden Fence

G - Safe Gambling

H - Government Help

【题意】经济危机，政府给予bankA和bankB补助，补助是按照一包一包的算。总共有N包，为了公平起见，怎么分配才能使得最后bankA和bankB得到的补助之差最小。

【思路】贪心策略，从小到大排序，先把最小的给A。然后从最大的开始先给B，再给A，依次分配下去，直到分完。代码如下：

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 
 5 using namespace std;
 6 
 7 int a[50005];
 8 int main() {
 9 #ifdef HotWhite
10    // freopen("in.txt", "r", stdin);
11 #endif
12     int n;
13     while(scanf("%d",&n))
14     {
15         if (!n) break;
16         for (int i = 0; i < n; i++)
17         {
18             scanf("%d",&a[i]);
19         }
20         sort(a,a+n);
21         printf("%d-A",a[0]);
22         int k=0;
23         for (int i = n-1; i >=1; i--,k++)
24         {
25             printf(" %d-%c",a[i],k&1?‘A‘:‘B‘);
26         }
27         printf("\n");
28     }
29     return 0;
30 }
View Code

 

I - Tree Insertions

 


联系
我们