1137. Final Grading (25)

1137. Final Grading (25)

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 G = (Gmid-termx 40% + Gfinalx 60%) if Gmid-term > Gfinal, or Gfinal will be taken as the final grade G. Here Gmid-term and Gfinal 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 Gp’s; the second one contains M mid-term scores Gmid-term’s; and the last one contains N final exam scores Gfinal’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 Gp Gmid-term Gfinal G

If some score does not exist, output “-1” 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 qualified 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 99ydjh2 200 98 82 88dx86w 220 88 81 84wehu8 300 55 84 84

解题思路：

定义结构体，node，其中GPout，Gout用于标记GP和G满不满足输出的条件，Gp,Gmid, Gfinal, G默认值都为-1。我先是用map把数据存起来，以名字为为key，node为value。最后把满足条件的数据放到vector<node> v中，用sort排序，输出即可

原文链接
代码如下：

#include<iostream>#include<algorithm>#include<string>#include<vector>#include<iterator>#include<map>using namespace std;struct node {    string ID;    int GPout=0,Gout=0;    int Gp=-1,Gmid=-1, Gfinal=-1, G=-1;};bool cmp(node a,node b) {    if (a.G != b.G)return a.G > b.G;    else return a.ID < b.ID;}int main(){    int P, M, N;    cin >> P >> M >> N;    map<string, node> m;    for (int i = 0; i < P; i++) {        string ID;        int GP;        cin >> ID >> GP;        m[ID].ID = ID;        m[ID].Gp = GP;        if (GP < 200)m[ID].GPout = 0;        else m[ID].GPout = 1;    }    for (int i = 0; i < M; i++) {        string ID;        int Gmid;        cin >> ID >> Gmid;        m[ID].ID = ID;        m[ID].Gmid = Gmid;    }    for (int i = 0; i < N; i++) {        string ID;        int Gfinal;        cin >> ID >> Gfinal;        m[ID].ID = ID;        m[ID].Gfinal = Gfinal;        if (m[ID].Gmid > m[ID].Gfinal) {            double d = (m[ID].Gmid*1.0*0.4 + m[ID].Gfinal*1.0*0.6)+0.5;            m[ID].G = (int)d;        }        else m[ID].G = m[ID].Gfinal;        if (m[ID].G >= 60 && m[ID].G <= 100)m[ID].Gout = 1;        else m[ID].Gout = 0·    }    vector<node> v;    for (map<string, node>::iterator it = m.begin(); it != m.end(); it++) {        if (it->second.GPout == 1 && it->second.Gout == 1) {            node temp;            v.push_back(it->second);        }    }    sort(v.begin(), v.end(), cmp);    for (int i = 0; i < v.size(); i++) {        printf("%s %d %d %d %d\n", v[i].ID.c_str(), v[i].Gp, v[i].Gmid, v[i].Gfinal, v[i].G);    }    return 0;}