PAT_A1137#Final Grading

Source:

PAT A1137 Final Grading (25 分)

Description:

For a student taking the online course “Data Structures” on China University MOOC (http://www.icourse163.org/), to be qualified for a certificate, he/she must first obtain no less than 200 points from the online programming assignments, and then receive a final grade no less than 60 out of 100. The final grade is calculated by 0 if G​mid−term​​>G​final​​, or G​final​​ will be taken as the final grade G. Here G​mid−term​​ and G​final​​ are the student’s scores of the mid-term and the final exams, respectively. The problem is that different exams have different grading sheets. Your job is to write a program to merge all the grading sheets into one.

Input Specification:

Each input file contains one test case. For each case, the first line gives three positive integers: P , the number of students having done the online programming assignments; M, the number of students on the mid-term list; and N, the number of students on the final exam list. All the numbers are no more than 10,000. Then three blocks follow. The first block contains P online programming scores G​p​​’s; the second one contains M mid-term scores G​mid−term​​’s; and the last one contains N final exam scores G​final​​’s. Each score occupies a line with the format: StudentID Score, where StudentID is a string of no more than 20 English letters and digits, and Score is a nonnegative integer (the maximum score of the online programming is 900, and that of the mid-term and final exams is 100).

Output Specification:

For each case, print the list of students who are qualified for certificates. Each student occupies a line with the format: StudentID G​p​​ G​mid−term​​ G​final​​ G If some score does not exist, output “−” instead. The output must be sorted in descending order of their final grades (G must be rounded up to an integer). If there is a tie, output in ascending order of their StudentID‘s. It is guaranteed that the StudentID‘s are all distinct, and there is at least one qullified student.

Sample Input:

6 6 7 01234 880 a1903 199 ydjh2 200 wehu8 300 dx86w 220 missing 400 ydhfu77 99 wehu8 55 ydjh2 98 dx86w 88 a1903 86 01234 39 ydhfu77 88 a1903 66 01234 58 wehu8 84 ydjh2 82 missing 99 dx86w 81

Sample Output:

missing 400 -1 99 99 ydjh2 200 98 82 88 dx86w 220 88 81 84 wehu8 300 55 84 84

Keys:

• 快乐模拟
• map（C++ STL）
• string（C++ STL）

Attention:

• 四舍五入要用round()函数；
• MAXSIZE最多有三倍的1E4；

Code:

  1 /*
  2 Data: 2019-05-26 21:12:36
  3 Problem: PAT_A1137#Final Grading
  4 AC: 55:00
  5 
  6 题目大意：
  7 获得证书需要，编程任务不少于200分，总分不少于60分
  8 如果期中>期末分数，则总分=期中*0.4+期末*0.6
  9 反之，总分=期末分数
 10 输入：
 11 第一行给出，完成编程的人数P，参加期中考试的人数M，参加期末考试的人数N，均<=1e4
 12 接下来给出各项分数；id不超过20位
 13 */
 14 
 15 #include<cstdio>
 16 #include<string>
 17 #include<camth>
 18 #include<map>
 19 #include<iostream>
 20 #include<algorithm>
 21 using namespace std;
 22 const int M=1e4+10;
 23 int pt=1,pass[M]={
   0};
 24 struct node
 25 {
 26     string id;
 27     int gp,gm,gf,g;
 28 }info[M],ans[M];
 29 map<string,int> mp;
 30 
 31 int ToInt(string s)
 32 {
 33     if(mp[s]==0)
 34     {
 35         mp[s]=pt;
 36         info[pt].id=s;
 37         info[pt].gm=-1;
 38         info[pt].gf=-1;
 39         return pt++;
 40     }
 41     else
 42         return mp[s];
 43 }
 44 
 45 bool cmp(node a, node b)
 46 {
 47     if(a.g != b.g)
 48         return a.g > b.g;
 49     else
 50         return a.id < b.id;
 51 }
 52 
 53 int main()
 54 {
 55 #ifdef    ONLINE_JUDGE
 56 #else
 57     freopen("Test.txt", "r", stdin);
 58 #endif
 59 
 60     int n,m,p,g;
 61     string s;
 62     scanf("%d%d%d", &n,&m,&p);
 63     for(int i=0; i<n; i++)
 64     {
 65         cin >> s >> g;
 66         int v=ToInt(s);
 67         info[v].gp=g;
 68         pass[v]++;
 69     }
 70     for(int i=0; i<m; i++)
 71     {
 72         cin >> s >> g;
 73         int v=ToInt(s);
 74         info[v].gm=g;
 75         if(pass[v])  pass[v]++;
 76     }
 77     for(int i=0; i<p; i++)
 78     {
 79         cin >> s >> g;
 80         int v=ToInt(s);
 81         info[v].gf=g;
 82         if(pass[v])  pass[v]++;
 83     }
 84     int cnt=0;
 85     for(int i=1; i<pt; i++)
 86     {
 87         if(pass[i]>=2 && info[i].gp>=200)
 88         {
 89             if(info[i].gm > info[i].gf)
 90                 info[i].g = (int)round(0.4*info[i].gm+0.6*info[i].gf);
 91             else
 92                 info[i].g = info[i].gf;
 93             if(info[i].g >= 60)
 94                 ans[cnt++] = info[i];
 95         }
 96     }
 97     sort(ans,ans+cnt,cmp);
 98     for(int i=0; i<cnt; i++)
 99     {
100         cout << ans[i].id;
101         printf(" %d %d %d %d\n", ans[i].gp,ans[i].gm,ans[i].gf,ans[i].g);
102     }
103 
104     return 0;
105 }

转载于:https://www.cnblogs.com/blue-lin/p/10932423.html

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