PAT 1137 Final Grading[一般][排序]

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 G=(Gmid−term×40%+Gfinal×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 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

题目大意：看完题感觉还是比较简单的，学生们都有一个id，不超过20个英文字母和数字，并且有上机成绩、期中成绩、期末成绩、最终成绩。最终成绩的判断是由期中成绩*40%+期末成绩*60%，如果期中成绩<期末成绩，那直接由期末成绩构成。那么谁能够获得证书呢？编程成绩>=200&&最终成绩>=60。

//首先我想用map，但是map可以映射一个结构体吗?关键是这个id怎么映射啊？！

//用map映射啊，你忘了那道图论的如何到ROM的题目了吗？就是把每个城市的名字映射成id,还有一个map是将id映射为名字。

#include <iostream>

#include <map>

#include <vector>

#include <algorithm>

#include<math.h>

using namespace std;

map<string,int> mp;

struct Stu{

    string id;

    int gp,gmid,gfin,g;

    Stu(){gp=-;gmid=-;gfin=-;

    }

}stu[];

bool cmp(Stu& a,Stu& b){

    return a.g!=b.g?a.g>b.g:a.id<b.id;

}

int main() {

    int m,n,k;

    cin>>m>>n>>k;

    int cnt=,temp;//共有多少个人。

    string s;

    for(int i=;i<m;i++){

        cin>>s>>temp;

        if(mp[s]==){

            mp[s]=cnt++;

            stu[mp[s]].id=s;

        }

        stu[mp[s]].gp=temp;

    }

    for(int i=;i<n;i++){

        cin>>s>>temp;

        if(mp[s]==){

            mp[s]=cnt++;

            stu[mp[s]].id=s;

        }

        stu[mp[s]].gmid=temp;

    }

    for(int i=;i<k;i++){

        cin>>s>>temp;

        if(mp[s]==){

            mp[s]=cnt++;

            stu[mp[s]].id=s;

        }

        stu[mp[s]].gfin=temp;

        if(stu[mp[s]].gmid<=stu[mp[s]].gfin){

            stu[mp[s]].g=stu[mp[s]].gfin;

        }else {

            stu[mp[s]].g=round(stu[mp[s]].gmid*0.4+stu[mp[s]].gfin*0.6);

            //怎么才能在这四舍五入呢？！

        }

    }

    sort(stu,stu+cnt,cmp);

    for(int i=;i<cnt;i++){

        if(stu[i].gp>=&&stu[i].g>=)

            cout<<stu[i].id<<" "<<stu[i].gp<<" "<<stu[i].gmid<<" "<<stu[i].gfin<<" "<<stu[i].g<<'\n';

    }

    return ;

}

//写成了这样，第一次提交编译错误，发现是round函数，在CB中是好使的，但是提交到pat平台上时需要加上math.h头文件！

//第二次提交，发生了答案错误和段错误，后两个测试点没过。

//我知道为什么发生段错误了，因为虽然m,n,k每一个数都是在10000之内，但是并不能保证总人数就是10000之内啊！

#include <iostream>

#include <map>

#include <vector>

#include <algorithm>

#include<math.h>

using namespace std;

map<string,int> mp;

struct Stu

{

    string id;

    int gp,gmid,gfin,g;

    Stu()

    {

        gp=-;

        gmid=-;

        gfin=-;

    }

} stu[];

bool cmp(Stu& a,Stu& b)

{

    return a.g!=b.g?a.g>b.g:a.id<b.id;

}

int main()

{

    int m,n,k;

    cin>>m>>n>>k;

    int cnt=,temp;//共有多少个人。

    string s;

    for(int i=; i<m; i++) //如果机试不够或者是没有机试分数，那必然不能获得证书。

    {

        cin>>s>>temp;

        if(temp>=)

        {

            mp[s]=cnt++;

            stu[mp[s]].id=s;

            stu[mp[s]].gp=temp;

        }

        else

        {

            mp[s]=-;

        }

    }

    for(int i=; i<n; i++)

    {

        cin>>s>>temp;

        if(mp[s]!=-) //而且机试成绩是满足条件的！

        {

            if(mp[s]==)

            {

                mp[s]=cnt++;

                stu[mp[s]].id=s;

            }

            stu[mp[s]].gmid=temp;

        }

    }

    for(int i=; i<k; i++)

    {

        cin>>s>>temp;

        if(mp[s]!=-)

        {

            if(mp[s]==)

            {

                mp[s]=cnt++;

                stu[mp[s]].id=s;

            }

            stu[mp[s]].gfin=temp;

            if(stu[mp[s]].gmid<=stu[mp[s]].gfin)

            {

                stu[mp[s]].g=stu[mp[s]].gfin;

            }

            else

            {

                stu[mp[s]].g=round(stu[mp[s]].gmid*0.4+stu[mp[s]].gfin*0.6);

                //怎么才能在这四舍五入呢？！

            }

        }

    }

    sort(stu,stu+cnt,cmp);

    for(int i=; i<cnt; i++)

    {

        if(stu[i].gp>=&&stu[i].g>=)

            cout<<stu[i].id<<" "<<stu[i].gp<<" "<<stu[i].gmid<<" "<<stu[i].gfin<<" "<<stu[i].g<<'\n';

    }

    return ;

}

//后来我这样去写，测试点2过不去，后来才发现，有可能是一个人的机试成绩，并没有在输入机试时存在，那么此时她就是-1！！。需要再改判断条件！

#include <iostream>

#include <map>

#include <vector>

#include <algorithm>

#include<math.h>

using namespace std;

map<string,int> mp;

struct Stu

{

    string id;

    int gp,gmid,gfin,g;

    Stu()

    {

        gp=-;

        gmid=-;

        gfin=;

        g=;//我的妈呀，我这里居然没有初始化g，难道是这里的错？

    }

} stu[];

bool cmp(Stu& a,Stu& b)

{

    return a.g!=b.g?a.g>b.g:a.id<b.id;

}

int main()

{

    int m,n,k;

    cin>>m>>n>>k;

    int cnt=,temp;//共有多少个人。

    string s;

    for(int i=; i<m; i++) //如果机试不够或者是没有机试分数，那必然不能获得证书。

    {

        cin>>s>>temp;

        if(temp>=)

        {

            mp[s]=cnt++;

            stu[mp[s]].id=s;

            stu[mp[s]].gp=temp;

        }

    }

    for(int i=; i<n; i++)

    {

        cin>>s>>temp;

            if(mp[s]!=)//其实只要统计机试中出现过的人就好！

            {

                stu[mp[s]].gmid=temp;

            }

    }

    for(int i=; i<k; i++)

    {

        cin>>s>>temp;

            if(mp[s]!=)//如果没有期末成绩的，那么一定不能输出，这就找到我的问题了，

                //我下面排序和输出都是用的cnt!!!.那么还需要再计一次！

            {

            stu[mp[s]].gfin=temp;

            if(stu[mp[s]].gmid<=stu[mp[s]].gfin)

            {

                stu[mp[s]].g=stu[mp[s]].gfin;

            }

            else

            {

                stu[mp[s]].g=int(stu[mp[s]].gmid*0.4+stu[mp[s]].gfin*0.6+0.5);

                //怎么才能在这四舍五入呢？！

            }

            }

    }

    //if(cnt!=ct)cnt=ct;不能这样，因为那些原来的不符合的仍然在stu里顺序存在呢。。。

    //看来只能在for循环一遍了，将答案存储再去排序和输出。

    vector<Stu> ans;

    for(int i=;i<cnt;i++){

        if(stu[i].g>=)ans.push_back(stu[i]);

    }

    sort(ans.begin(),ans.end(),cmp);

    for(int i=; i<ans.size(); i++)

    {

        cout<<ans[i].id<<" "<<ans[i].gp<<" "<<ans[i].gmid<<" "<<ans[i].gfin<<" "<<ans[i].g<<'\n';

    }

    return ;

}

//终于AC了，终于想出来原因了。智障吧，你用mp[s]==0去判断这个s是否存在，你还居然一开始把cnt初始化为0，让第一个人映射到0！！！！你这不是智障是啥。