1012. The Best Rank (25)——PAT (Advanced Level) Practise
来源:互联网 发布:mt4软件下载 编辑:程序博客网 时间:2024/05/17 19:17
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
代码如下:
#include <iostream>#include <string>#include <map>#include <set>using namespace std;class CJ{public:int c, m, e, a;int cp, mp, ep, ap;CJ(){cp = mp = ep = ap = 0;}};int main(){map<string, CJ> mp;map<int, string> mpb;int n, m;while (!cin.eof() && cin >> n >> m){for (int i = 0; i < n; ++i){string str;cin >> str;CJ cj;cin >> cj.c >> cj.m >> cj.e;cj.a = (cj.c + cj.m + cj.e) / 3;mp[str] = cj;}for (int j = 0; j < m; ++j){string str;cin >> str;mpb[j] = str;}for (map<string, CJ>::iterator ipi = mp.begin();ipi != mp.end(); ipi++){for (map<string, CJ>::iterator ipj = ipi;ipj != mp.end(); ipj++){if (ipi == ipj)continue;if (ipi->second.c < ipj->second.c)ipi->second.cp++;else if (ipi->second.c > ipj->second.c)ipj->second.cp++;if (ipi->second.m < ipj->second.m)ipi->second.mp++;else if (ipi->second.m > ipj->second.m)ipj->second.mp++;if (ipi->second.e < ipj->second.e)ipi->second.ep++;else if (ipi->second.e > ipj->second.e)ipj->second.ep++;if (ipi->second.a < ipj->second.a)ipi->second.ap++;else if (ipi->second.a > ipj->second.a)ipj->second.ap++;}}for (map<int, string>::iterator ip = mpb.begin();ip != mpb.end(); ip++){if (mp.find(ip->second) != mp.end()){int arr[4] = { mp[ip->second].ap, mp[ip->second].cp, mp[ip->second].mp, mp[ip->second].ep };int max = 0;for (int i = 1; i < 4; i++){if (arr[i] < arr[max])max = i;}cout << arr[max] + 1 << " ";char ch;switch (max){case 0:ch = 'A';break;case 1:ch = 'C';break;case 2:ch = 'M';break;case 3:ch = 'E';}cout << ch << endl;}elsecout << "N/A" << endl;}}return 0;}
- PAT (Advanced Level) Practise 1012. The Best Rank (25)
- PAT (Advanced Level) Practise 1012. The Best Rank (25)
- 1012. The Best Rank (25)——PAT (Advanced Level) Practise
- 1012. The Best Rank @ PAT (Advanced Level) Practise
- 浙大PAT (Advanced Level) Practise 1012 The Best Rank (25)
- PAT (Advanced Level) Practise 1012 The Best Rank (25)
- PAT (Advanced Level) Practise 1012 The Best Rank
- PAT-PAT (Advanced Level) Practise 1012 The Best Rank (25) (简单模拟)【二星级】
- 【PAT Advanced Level】1012. The Best Rank (25)
- 【c++】PAT (Advanced Level)1012. The Best Rank (25)*
- 【PAT】【Advanced Level】1012. The Best Rank (25)
- PAT (Advanced) 1012. The Best Rank (25)
- PAT (Advanced) 1012. The Best Rank (25)
- [PAT (Advanced Level) ]1012. The Best Rank 解题文档
- 浙大 PAT Advanced level 1012. The Best Rank
- PAT (Advanced Level) 1012. The Best Rank (25) 最佳排名,结构体排序,哈希查询
- PAT(Advanced lecel) 1012. The Best Rank
- Pat(Advanced Level)Practice--1012(The Best Rank)
- iPhone开发笔记(11)用ASIFormDataRequest实现图片上传
- c#定时期调用方法
- Android数据库高手秘籍(四)——使用LitePal建立表关联
- golang 组合算法
- iPhone开发笔记(9)ASIHttpRequest和json-framework实现json解析(iOS客户端)
- 1012. The Best Rank (25)——PAT (Advanced Level) Practise
- Eclipse快捷键
- ——谈VC++对象模型
- 【android】网络定位服务NetworkLocationProvider
- 好好学习Linux
- Android Loaders介绍(异步加载数据利器,类似AsyncTask)
- Dialog的自定义view无法获取到点击事件
- u盘raw数据恢复,小牛数据恢复软件
- 45个常用的Oracle查询语句