PAT甲级-1012. The Best Rank (25)并列排序

心已赠人 2022-05-31 09:57 133阅读 0赞

# 1012. The Best Rank (25) #

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

To evaluate the performance of our first year CS majored students, we consider their grades of three courses only: C - C Programming Language, M - Mathematics (Calculus or Linear Algebra), and E - English. At the mean time, we encourage students by emphasizing on their best ranks -- that is, among the four ranks with respect to the three courses and the average grade, we print the best rank for each student.

For example, The grades of C, M, E and A - Average of 4 students are given as the following:

StudentID  C  M  E  A
    310101     98 85 88 90
    310102     70 95 88 84
    310103     82 87 94 88
    310104     91 91 91 91

Then the best ranks for all the students are *No.1* since the 1st one has done the best in C Programming Language, while the 2nd one in Mathematics, the 3rd one in English, and the last one in average.

**Input**

Each input file contains one test case. Each case starts with a line containing 2 numbers N and M (<=2000), which are the total number of students, and the number of students who would check their ranks, respectively. Then N lines follow, each contains a student ID which is a string of 6 digits, followed by the three integer grades (in the range of \[0, 100\]) of that student in the order of C, M and E. Then there are M lines, each containing a student ID.

**Output**

For each of the M students, print in one line the best rank for him/her, and the symbol of the corresponding rank, separated by a space.

The priorities of the ranking methods are ordered as A > C > M > E. Hence if there are two or more ways for a student to obtain the same best rank, output the one with the highest priority.

If a student is not on the grading list, simply output "N/A".

**Sample Input**

5 6
    310101 98 85 88
    310102 70 95 88
    310103 82 87 94
    310104 91 91 91
    310105 85 90 90
    310101
    310102
    310103
    310104
    310105
    999999

**Sample Output**

1 C
    1 M
    1 E
    1 A
    3 A
    N/A

给出N个学生的ID及其编程、数学、英语的分数，计算其平均分，排序后根据给定ID查询得出学生的各科成绩中排名最好的单科成绩，输出排名及科目名称。

注意排名含有相同分数并列的情况时，采取名次并列，后继递减的方法，例如第一第二并列第一时，第三名还是第三，即 1 1 3 4。

#include<bits/stdc++.h>
    using namespace std;
    #define INF 0xfffffff
    #define MAXN 2010
    struct N
    {
        string id;
        double ave,c,m,e;
        int no,na,nc,nm,ne;//最好排名，均分排名，编程排名，数学排名，英语排名
        char sub;//排名最好的科目的名称
    } a[MAXN];
    bool cmp1(N x,N y)//按均分排
    {
        return x.ave>y.ave;
    }
    bool cmp2(N x,N y)//按编程分排
    {
        return x.c>y.c;
    }
    bool cmp3(N x,N y)//按数学分排
    {
        return x.m>y.m;
    }
    bool cmp4(N x,N y)//按英语分排
    {
        return x.e>y.e;
    }
    int main()
    {
    #ifdef ONLINE_JUDGE
    #else
        freopen("F:/cb/read.txt","r",stdin);
    //freopen("F:/cb/out.txt","w",stdout);
    #endif
        ios::sync_with_stdio(false);
        cin.tie(0);
        int n,q;
        cin>>n>>q;
        for(int i=0; i<n; ++i)
        {
            cin>>a[i].id>>a[i].c>>a[i].m>>a[i].e;
            a[i].ave=(a[i].c+a[i].m+a[i].e)/3.0;
        }
        bool flag=false;
        int no=INF;
        sort(a,a+n,cmp1);
        int cnt=0;//记录相同分数的学生人数
        a[0].na=cnt;
        for(int i=1; i<n; ++i)
        {
            if(a[i].ave==a[i-1].ave) ++cnt;//筛重
            else cnt=0;
            a[i].na=i-cnt;
        }
        sort(a,a+n,cmp2);
        cnt=0;
        a[0].nc=cnt;
        for(int i=1; i<n; ++i)
        {
            if(a[i].c==a[i-1].c) ++cnt;
            else cnt=0;
            a[i].nc=i-cnt;
        }
        sort(a,a+n,cmp3);
        cnt=0;
        a[0].nm=cnt;
        for(int i=1; i<n; ++i)
        {
            if(a[i].m==a[i-1].m) ++cnt;
            else cnt=0;
            a[i].nm=i-cnt;
        }
        sort(a,a+n,cmp4);
        cnt=0;
        a[0].ne=cnt;
        for(int i=1; i<n; ++i)
        {
            if(a[i].e==a[i-1].e) ++cnt;
            else cnt=0;
            a[i].ne=i-cnt;
        }
        for(int i=0; i<n; ++i)//从各科排名中选出最好排名
        {
            if(a[i].na<=a[i].nc&&a[i].na<=a[i].nm&&a[i].na<=a[i].ne)      a[i].no=a[i].na,a[i].sub='A';
            else if(a[i].nc<=a[i].na&&a[i].nc<=a[i].nm&&a[i].nc<=a[i].ne) a[i].no=a[i].nc,a[i].sub='C';
            else if(a[i].nm<=a[i].nc&&a[i].nm<=a[i].na&&a[i].nm<=a[i].ne) a[i].no=a[i].nm,a[i].sub='M';
            else if(a[i].ne<=a[i].nc&&a[i].ne<=a[i].nm&&a[i].ne<=a[i].na) a[i].no=a[i].ne,a[i].sub='E';
        }
        for(int i=0; i<q; ++i)
        {
            string s;
            cin>>s;
            flag=false;
            for(int j=0; j<n; ++j)
                if(s==a[j].id)
                {
                    flag=true;
                    no=j;
                    break;
                }
            if(flag) cout<<++a[no].no<<" "<<a[no].sub<<endl;
            else cout<<"N/A"<<endl;
        }
    }