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−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：在慕课上学习数据结构的同学想要获得证书，需要满足一些条件。首先程序设计考试得分要大于等于200分，其次需要总评成绩大于等于60小于等于100。总评成绩根据期中和期末考试成绩得到，如果期中考试成绩大于期末考试，则总评成绩等于向上取整的40%期中+60%期末分数；如果期中考试成绩小于期末考试，则总评成绩等于期末成绩。

现给出p条程序考试记录，m条期中考试记录，n条期末考试记录。每条记录由考生姓名、对应成绩组成。要求按照成绩降序，成绩相同则按照名字升序输出所有总评分数大于等于60小于等于100的学生，以及程序考试、期中考试、期末考试成绩，如果某项成绩不存在，对应位置输出-1。

分析：问题主要是如何查找到已经出现过的学生。可以用3个数组分别存储程序考试、期中考试、期末考试的记录，之后用三个指针同时遍历三个数组进行匹配；也可以再读入后两个考试的记录时，二分查找程序考试的记录。

#include<algorithm>
#include <iostream>
#include  <cstdlib>
#include  <cstring>
#include   <string>
#include   <vector>
#include   <cstdio>
#include    <queue>
#include    <stack>
#include    <ctime>
#include    <cmath>
#include      <map>
#include      <set>
#define INF 0xffffffff
#define db1(x) cout<<#x<<"="<<(x)<<endl
#define db2(x,y) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<endl
#define db3(x,y,z) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<endl
#define db4(x,y,z,r) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<", "<<#r<<"="<<(r)<<endl
#define db5(x,y,z,r,w) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<", "<<#r<<"="<<(r)<<", "<<#w<<"="<<(w)<<endl
using namespace std;

typedef struct node
{
    string name;
    int sp,sm,sn,se;
}node;

bool cmp(node a,node b)
{
    if(a.se!=b.se)return a.se>b.se;
    return a.name<b.name;
}

bool cmpname(node a,node b)
{
    return a.name<b.name;
}

int main(void)
{
    #ifdef test
    freopen("in.txt","r",stdin);
    //freopen("in.txt","w",stdout);
    clock_t start=clock();
    #endif //test

    int p,m,n,t=0;scanf("%d%d%d",&p,&m,&n);
    node num[10010];
    for(int i=0;i<10010;++i)
        num[i].sm=num[i].se=-1;
    for(int i=0;i<p;++i)
    {
        string n;int a;cin>>n>>a;
        if(a>=200)num[t].name=n,num[t].sp=a,t++;
    }sort(num,num+t,cmpname);
    int nsm=0;
    node numsm[10010];
    for(int i=0;i<m;++i)
    {
        string n;int a;cin>>n>>a;
        numsm[nsm].name=n,numsm[nsm].sm=a,nsm++;
    }sort(numsm,numsm+nsm,cmpname);
    int nsn=0;
    node numsn[10010];
    for(int i=0;i<n;++i)
    {
        string n;int a;cin>>n>>a;
        numsn[nsn].name=n,numsn[nsn].sn=a,nsn++;
    }sort(numsn,numsn+nsn,cmpname);

//    for(int i=0;i<t;++i)
//        db2(num[i].name,num[i].sp);
//    printf("\n");
//    for(int i=0;i<nsm;++i)
//        db2(numsm[i].name,numsm[i].sm);
//    printf("\n");
//    for(int i=0;i<nsn;++i)
//        db2(numsn[i].name,numsn[i].sn);
//    printf("\n");

    int indexsm=0,indexsn=0;
    for(int i=0;i<t;++i)
    {
//        db2(i,num[i].name);
        while(numsm[indexsm].name<=num[i].name&&indexsm<nsm)
        {
            if(num[i].name==numsm[indexsm].name)num[i].sm=numsm[indexsm].sm;
            indexsm++;
        }
        while(numsn[indexsn].name<=num[i].name&&indexsn<nsn)
        {
            if(num[i].name==numsn[indexsn].name)num[i].sn=numsn[indexsn].sn;
            indexsn++;
        }
    }

    for(int i=0;i<t;++i)
    {
        if(num[i].sm>num[i].sn)
        {
            int temp=(num[i].sm*0.4+num[i].sn*0.6)*10;
            if(temp%10<=4)num[i].se=temp/10;
            else num[i].se=temp/10+1;
        }
        else num[i].se=max(0,num[i].sn);
//        db5(num[i].name,num[i].sp,num[i].sm,num[i].sn,num[i].se);
    }
    sort(num,num+t,cmp);
    for(int i=0;i<t;++i)
    {
        if(num[i].se>=60)
        {
            cout<<num[i].name<<' '<<num[i].sp<<' '<<num[i].sm<<' '<<num[i].sn<<' '<<num[i].se<<endl;
        }
    }

    #ifdef test
    clockid_t end=clock();
    double endtime=(double)(end-start)/CLOCKS_PER_SEC;
    printf("\n\n\n\n\n");
    cout<<"Total time:"<<endtime<<"s"<<endl;        //s为单位
    cout<<"Total time:"<<endtime*1000<<"ms"<<endl;    //ms为单位
    #endif //test
    return 0;
}