pat 1137

1137 Final Grading (25分)

 

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

题意:给定三组数据,分别是学生的编程成绩、期中成绩和期末成绩。要求按最终分数从高到低输出及格的学生,分数重复的按照名字排序。其中最终分数=期中分数*0.4+期末分数*0.6(如果期中分数大于期末),否则直接将期末成绩作为最终成绩。编程分数不足200的学生不计入统计。最后分数需要四舍五入到整数位。

思路:自定义结点node,node包括编程分数、期中分数和期末分数。输入的时候如果编程分数>200则将这个学生的记录插入map<string,node>,其中string记录学生名字。当输入期中和期末成绩的时候,如果这个学生的姓名不存在map中,直接忽略。遍历map,计算总成绩,将总成绩大于等于60的加入vector中,最后对vector排序输出。

代码如下

#include<cstdio>
#include<iostream>
#include<vector>
#include<map>
#include<string>
#include<cstring>
#include<algorithm>
#include<math.h>
using namespace std;
struct node2{
    string name;
    int sP;
    int sM;
    int sN;
    int G;
}; 
struct node{
    int sP;
    int sM;
    int sN;
};
map<string,node> maps;
bool cmp(node2 a,node2 b){
    if(a.G!=b.G)
        return a.G>b.G;
    else
        return a.name<b.name;
}
int main(){
    int p,m,n;
    scanf("%d%d%d",&p,&m,&n);
    char name[200];
    int sP,sM,sN;
    for(int i=0;i<p;i++){
        scanf("%s %d",name,&sP);
        node temp;
        temp.sP=sP;
        temp.sM=temp.sN=-1;
        if(temp.sP>=200)
            maps.insert(make_pair(name,temp));

    }
    for(int i=0;i<m;i++){
        scanf("%s %d",name,&sM);
            if(maps.count(name)==1){
                maps[name].sM=sM;
            }
    }
    for(int i=0;i<n;i++){
        scanf("%s %d",name,&sN);
            if(maps.count(name)==1){
                maps[name].sN=sN;
            }    
    }
    vector<node2> v;
    for(map<string,node>::iterator it=maps.begin();it!=maps.end();it++){
            int total;
            if((*it).second.sM>(*it).second.sN){
                 total=round((*it).second.sM*0.4+(*it).second.sN*0.6);
            }
            else{
                 total=(*it).second.sN;
            }
            if(total>=60){
                node2 temp;
                temp.G=total;
                temp.sP=(*it).second.sP;
                temp.sM=(*it).second.sM;
                temp.sN=(*it).second.sN;
                temp.name=(*it).first;
                v.push_back(temp);
            }
        
    }
    sort(v.begin(),v.end(),cmp);
    for(int i=0;i<v.size();i++){
        printf("%s %d %d %d %d\n",v[i].name.c_str(),v[i].sP,v[i].sM,v[i].sN,v[i].G);
    }
    
} 

 

posted @ 2020-07-13 19:37  9761滴  阅读(4)  评论(0编辑  收藏