# 1012. The Best Rank (25)

400 ms

65536 kB

16000 B

Standard

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```
```#include<iostream>
#include<string>
using namespace std;
struct Student
{
string id;
int average;
int cRank;
int mRank;
int eRank;
int aRank;
}stu,sstu;
int Average(const int  c,const int m,const int e)
{
return (c+m+e)/3;
}
int max(const int cr,const int mr,const int er,const int ar,char &ch)
{
int temp=ar<(er<(cr<mr?cr:mr)?er:(cr<mr?cr:mr))?ar:(er<(cr<mr?cr:mr)?er:(cr<mr?cr:mr));
if (temp==ar)
ch='A';
else if(temp==cr)
ch='C';
else if (temp==mr)
ch='M';
else if (temp==er)
ch='E';
return temp;
}
int main()
{
int num;//输入学生数量
int snum;//查找学生编号数量
cin>>num>>snum;
for(int i=0;i<num;i++)
{
}
for(int i=0;i<num;i++)
{
stu[i].cRank=1;
stu[i].mRank=1;
stu[i].aRank=1;
stu[i].eRank=1;
for(int j=0;j<num;j++)
{
stu[i].cRank++;
stu[i].mRank++;
stu[i].eRank++;
if(stu[i].average<stu[j].average)
stu[i].aRank++;
}
}
for(int i=0;i<snum;i++)
cin>>sstu[i].id;
for(int i=0;i<snum;i++)
{
bool flag=false;
for(int j=0;j<num;j++)
{
if(sstu[i].id==stu[j].id)
{
flag=true;
char ch;
int temp=max(stu[j].cRank,stu[j].mRank,stu[j].eRank,stu[j].aRank,ch);
cout<<temp<<" "<<ch<<endl;
break;
}
}
if(flag==false)
cout <<"N/A"<<endl;
}
return 0;
}```

494 篇文章45 人订阅

0 条评论

## 相关文章

### 2018-04-17 Java的Collection集合类3分钟搞掂Set集合前言

3分钟搞掂Set集合 前言 声明，本文用的是jdk1.8 现在这篇主要讲Set集合的三个子类： HashSet集合 A:底层数据结构是哈希表(是一个元素为链...

30770

### 聊聊storm的WindowedBolt

storm-2.0.0/storm-client/src/jvm/org/apache/storm/topology/IWindowedBolt.java

8920

### hdu 4033Regular Polygon(二分+余弦定理)

Regular Polygon Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65...

33960

34740

25710

12450

11520

### BZOJ 2456: mode(新生必做的水题)

2456: mode Time Limit: 1 Sec  Memory Limit: 1 MB Submit: 4868  Solved: 2039 Des...

27660

### 3402: [Usaco2009 Open]Hide and Seek 捉迷藏

3402: [Usaco2009 Open]Hide and Seek 捉迷藏 Time Limit: 3 Sec  Memory Limit: 128 MB ...

34770

### 聊聊storm的reportError 