PAT-A-1012. The Best Rank (25)
来源:互联网 发布:拍拍网和淘宝网哪个好 编辑:程序博客网 时间:2024/04/29 15:18
1012. The Best Rank (25)
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 A310101 98 85 88 90310102 70 95 88 84310103 82 87 94 88310104 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 Input5 6310101 98 85 88310102 70 95 88310103 82 87 94310104 91 91 91310105 85 90 90310101310102310103310104310105999999Sample Output
1 C1 M1 E1 A3 AN/A
提交代
码
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int D[1000000][4] = { 0 };
int sub=0;
struct student
{
int id;
int g[4];
}stu[2010];
int cmp(student s1, student s2)
{
return s1.g[sub] > s2.g[sub];
}
int main()
{
char r[4] = { 'A', 'C', 'M', 'E' };
int n, m;
cin >> n >> m;
for (int i = 0; i < n; i++)
{
cin >> stu[i].id >> stu[i].g[1] >> stu[i].g[2] >> stu[i].g[3];
stu[i].g[0] = (stu[i].g[1] + stu[i].g[2] + stu[i].g[3]) / 3 + 0.5;
}
for (sub = 0; sub < 4; sub++)
{
sort(stu, stu + n, cmp);
D[stu[0].id][sub] = 1;
for (int i = 1; i < n; i++)
{
if (stu[i].g[sub] == stu[i - 1].g[sub])
D[stu[i].id][sub] = D[stu[i - 1].id][sub];
else
D[stu[i].id][sub] = i + 1;
}
}
int s;
for (int i = 0; i < m; i++)
{
cin >> s;
if (D[s][0] == 0)
cout << "N/A" << endl;
else
{
int k = 0;
for (int j = 1; j < 4; j++)
{
if (D[s][j] < D[s][k])
k = j;
}
cout << D[s][k] << " " << r[k] << endl;
}
}
system("pause");
return 0;
}
- PAT A 1012. The Best Rank (25)
- PAT(A) - 1012. The Best Rank (25)
- PAT-A 1012. The Best Rank (25)
- PAT-A-1012. The Best Rank (25)
- PAT-A-1012. The Best Rank (25)
- PAT A 1012. The Best Rank
- 1012. The Best Rank (25)-PAT
- pat 1012. The Best Rank (25)
- 【PAT】1012. The Best Rank (25)
- PAT (Advanced) 1012. The Best Rank (25)
- PAT 1012. The Best Rank (25)
- PAT 1012. The Best Rank (25)
- PAT 1012. The Best Rank (25)
- PAT 1012. The Best Rank (25)
- PAT 1012. The Best Rank (25)
- PAT 1012. The Best Rank (25)
- PAT 1012. The Best Rank (25)
- [PAT]1012. The Best Rank (25)
- CTF解密图片得到flag
- EMVTag系列2《磁条等效数据》
- 创建Android虚拟设备(AVD)失败
- 工作中经常用到github上优秀、实用、轻量级、无依赖的插件和库
- 【cqoi】2017省选凑数记
- PAT-A-1012. The Best Rank (25)
- js模块化进程
- 【Maven】setting.xml中mirror和repository的关系
- 什么是JavaBean、bean? 什么是POJO、PO、DTO、VO、BO ? 什么是EJB、EntityBean?
- 关于using namespace std;
- EMV/PBOC 解析(二) 卡片数据读取
- Django 报错 LookupError: No installed app with label 'user' 的解决办法
- 226. Invert Binary Tree
- 正则表达式功能1