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 Algrbra), 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 Specification:

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 Specification:

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

题目大意:找出每个学生排名最高的科目,如果排名相同那么就A>C>M>E

思路分析:预处理一个排名查询表,输入分数即可查出排名

#include<bits/stdc++.h>
#define de(x) cout<<#x<<" "<<(x)<<endl
#define each(a,b,c) for(int a=b;a<=c;a++)
using namespace std;
const int maxn=2000+5;
const int inf=0x3f3f3f3f; int avg[maxn];
int c[maxn];
int math[maxn];
int english[maxn]; int avg_table[105];
int c_table[105];
int math_table[105];
int english_table[105];
struct student
{
int c;
int m;
int e;
int a;
};
map<string,student>M;
int cmp(int a,int b)
{
return a>b;
}
/*
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
*/
int main()
{
int n,m;
cin>>n>>m;
string id; each(i,1,n)
{
student temp;
cin>>id;
cin>>temp.c;
cin>>temp.m;
cin>>temp.e;
temp.a=(temp.c+temp.m+temp.e)/3;
c[i]=temp.c;
math[i]=temp.m;
english[i]=temp.e;
avg[i]=temp.a;
M[id]=temp;
}
sort(avg+1,avg+n+1,cmp);
for(int i=1;i<=n+1;i++)
{
int score=avg[i];
if(avg_table[score]==0)
{
avg_table[score]=i;
}
} sort(math+1,math+n+1,cmp);
for(int i=1;i<=n+1;i++)
{
int score=math[i];
if(math_table[score]==0)
{
math_table[score]=i;
}
} sort(c+1,c+n+1,cmp);
for(int i=1;i<=n+1;i++)
{
int score=c[i];
if(c_table[score]==0)
{
c_table[score]=i;
}
} sort(english+1,english+n+1,cmp);
for(int i=1;i<=n+1;i++)
{
int score=english[i];
if(english_table[score]==0)
{
english_table[score]=i;
}
} while(m--)
{
string query;
cin>>query;
if(M.count(query)==0)
{
cout<<"N/A"<<endl;
continue;
}
string ans1="E";
int ans2=english_table[M[query].e];
//de(ans2); if(math_table[M[query].m]<=english_table[M[query].e])
{
ans1="M";
ans2=math_table[M[query].m];
}
if(c_table[M[query].c]<=ans2)
{
ans1="C";
ans2=c_table[M[query].c];
}
if(avg_table[M[query].a]<=ans2)
{
ans1="A";
ans2=avg_table[M[query].a];
}
cout<<ans2<<" "<<ans1<<endl; }
}

PAT-1012 The Best Rank (25 分) 查询分数对应排名(包括并列)的更多相关文章

  1. 1012 The Best Rank (25分) vector与结构体排序

    1012 The Best Rank (25分)   To evaluate the performance of our first year CS majored students, we con ...

  2. PAT 甲级 1012 The Best Rank (25 分)(结构体排序)

    题意: 为了评估我们第一年的CS专业学生的表现,我们只考虑他们的三个课程的成绩:C - C编程语言,M - 数学(微积分或线性代数)和E - 英语.同时,我们鼓励学生强调自己的最优秀队伍 - 也就是说 ...

  3. 【PAT甲级】1012 The Best Rank (25 分)

    题意: 输入两个整数N,M(<=2000),接着分别输入N个学生的ID,C语言成绩,数学成绩和英语成绩. M次询问,每次输入学生ID,如果该ID不存在则输出N/A,存在则输出该学生排名最考前的一 ...

  4. 1012 The Best Rank (25 分)

    To evaluate the performance of our first year CS majored students, we consider their grades of three ...

  5. PAT甲 1012. The Best Rank (25) 2016-09-09 23:09 28人阅读 评论(0) 收藏

    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 eval ...

  6. [PAT] 1142 Maximal Clique(25 分)

    1142 Maximal Clique(25 分) A clique is a subset of vertices of an undirected graph such that every tw ...

  7. PAT 1009 Product of Polynomials (25分) 指数做数组下标,系数做值

    题目 This time, you are supposed to find A×B where A and B are two polynomials. Input Specification: E ...

  8. PAT A1122 Hamiltonian Cycle (25 分)——图遍历

    The "Hamilton cycle problem" is to find a simple cycle that contains every vertex in a gra ...

  9. PAT A1142 Maximal Clique (25 分)——图

    A clique is a subset of vertices of an undirected graph such that every two distinct vertices in the ...

随机推荐

  1. DELPHI搭建centos开发环境

    DELPHI搭建centos7开发环境 关闭防火墙 搭建开发环境,还是直接关闭LINUX防火墙,省事. 否则,使用到的网络端口号,都要在防火墙开放,麻烦. systemctl disable fire ...

  2. Hive中导入Oracle数据错误:Listener refused the connection with the following error: ORA-12505

    问题: 今天往Hive中导入Oracle数据的时候碰到了如下错误:Listener refused the connection with the following error: ORA-12505 ...

  3. CardView 简介和使用

    CardView 简介 本文链接:https://blog.csdn.net/ShawnXiaFei/article/details/81568537CardView 是 Google 官方发布 MD ...

  4. Win10系统安装VMware-viclient-6.0无响应问题解决方法

    背景:笔记本重做系统升级至Win10系统后,由于工作需要,得安装VMware-viclient-6.0软件进行远程连接. 问题:没有出现网上那种各种报错情况,只是在点击“安装”按钮的时候没弹出任何等待 ...

  5. SpringMVC源码分析--HandlerMappings

    之前分析过SpringMVC中的DispatcherServlet,分析了SpringMVC处理请求的过程.但忽略了一些DispatcherServlet协助请求处理的组件,例如SpringMVC中的 ...

  6. Linux下设置Tomcat开机自启动

    --未验证 第一步:在/etc/init.d下新建一个文件tomcat(需要root操作权限) vi /etc/init.d/tomcat 然后点击"i"写下如下代码,tomcat ...

  7. Windows 操作系统 端口转发

    在Windows 下可以使用netsh interface portproxy 命令实现端口转发功能. 例:netsh interface portproxy add v4tov4 listenpor ...

  8. Qt编写控件属性设计器7-串口采集

    一.前言 数据源是组态软件的核心灵魂,少了数据源,组态就是个花架子没卵用,一般数据源有三种方式获取,串口.网络.数据库,至于数据规则是什么,这个用户自己指定,本设计器全部采用第一个字节作为数据来演示. ...

  9. markdown 测试博客发布

    这是一个测试页面 无序列表 tet test 有序列表 特使团 tetst 引用 This is a test 插入图片 插入链接 baidu 粗体 这是粗体 斜体 这是斜体 表格 IP VIP 备注 ...

  10. (十一)使用Jconsole监控线程

    一.案例 监控线程情况,包括阻塞.死循环等 1.1 代码如下,下述代码共有三个线程,Main.mythread01.mythread02线程,其中mythread01线程为死循环.mythread02 ...