PTA (Advanced Level) 1012 The Best Rank
The Best Rank
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 (≤), 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
解题思路:
本题给出学生数量n,与查询数量m,之后n行为学生信息,包括id,C语言得分, 数学得分,英语得分,按照平均分、C语言、数学、英语分别排序(项目名分别为A, C, M, E),之后m行查询为学生id,要求输出查询学生的这四个排名中的最高排名,若出现某几项排名相同且都该学生最高排名,输出优先级平均分 > C语言 > 数学 > 英语。
定义一个结构体student记录每个学生的信息
struct student{
int id; //记录学号
int score[]; //score[0]为平均分 score[1]C语言 score[2]数学 score[3]英语
//为了方便输出直接按优先级顺序记录学生分数
};
用一个vector容器储存所有学生信息,一个int型的二维数组rank_stu[ i ][ j ]记录排名信息,i为学生排名, j为排名依照项,同样 0 为平均分 1 C语言 2 数学 3 英语 。之后对所有项目进行排名,将得到的排名记录入rank_stu中就可以开始查找了。
查找时要先判断输入的查询学号存不存在,可以将rank_stu初始化为0,如果查询当前学号某一排名为 0 证明该学号不存在,若存在,比较其所有项目排名并输出最高项排名与项目名。
AC代码
#include<bits/stdc++.h>
using namespace std;
int inf = INT_MAX;
//无穷大
const int maxn = ;
//学号最大值
const char course[] = {'A', 'C', 'M', 'E'};
//记录项目名
struct student{
int id; //记录学号
int score[]; //score[0]为平均分 score[1]C语言 score[2]数学 score[3]英语
//为了方便输出直接按优先级顺序记录学生分数
};
vector<student> stu; //记录学生信息
int rank_stu[maxn][]; //记录排名
int courseNow; //当前正在排名的课程
int n, m; //学生数量,查询数量
bool cmp(student a, student b){
return a.score[courseNow] > b.score[courseNow];
}
//排序依照当前courseNow的对应成绩
void getRank(){
memset(rank_stu, , sizeof(rank_stu));
//初始化rank_stu为0
for(int i = ; i < ; i++){ //获得并记录四个项目的排名
courseNow = i; //设定当前排名课程
sort(stu.begin(), stu.end(), cmp); //对stu进行排序
int cnt = ;
vector<student>::iterator preit = stu.begin(); //preit记录获取排名过程中上一个位置迭代器
for(vector<student>::iterator it = stu.begin(); it != stu.end(); it++){
cnt++; //记录当前学生位置
if(it == stu.begin()){ //若it排名第一直接记录排名为cnt(cnt当前值为1)
rank_stu[it->id][courseNow] = cnt;
preit = it; //记录下一个学生的前一个迭代器为it
}else{
if(preit->score[courseNow] == it->score[courseNow]){
//前一个学生与本学生成绩相同则排名相同
rank_stu[it->id][courseNow] = rank_stu[preit->id][courseNow];
}else{
rank_stu[it->id][courseNow] = cnt;
//否则记录当前排名为cnt
}
preit = it;//记录下一个学生的前一个迭代器为it
}
}
}
}
int main(){
while(scanf("%d%d", &n, &m) != EOF){ //输入学生数量与排名数量
while(n--){
student temp;
scanf("%d%d%d%d", &temp.id, &temp.score[], &temp.score[], &temp.score[]);
//输入学生信息
temp.score[] = (temp.score[] + temp.score[] + temp.score[]) / ;
//计算平均分
stu.push_back(temp);
} getRank();
//获取排名
while(m--){
int query;
scanf("%d", &query);
//输入查询id
if(rank_stu[query][] == ){ //学生不存在
printf("N/A\n");
continue;
}
int hRank = inf; //初始化最高排名为无穷大
int hRcourse = -; //hRcourse记录最高排名对应的项目
for(int i = ; i < ; i++){
if(rank_stu[query][i] < hRank){
hRank = rank_stu[query][i];
hRcourse = i;
}
}
printf("%d %c\n", hRank, course[hRcourse]);
//输出排名与项目名
}
}
return ;
}
PTA (Advanced Level) 1012 The Best Rank的更多相关文章
- PAT (Advanced Level) 1012. The Best Rank (25)
简单排序题. 注意:分数相同的人排名相同. #include<iostream> #include<cstring> #include<cmath> #includ ...
- PTA(Advanced Level)1036.Boys vs Girls
This time you are asked to tell the difference between the lowest grade of all the male students and ...
- PTA(Advanced Level)1025.PAT Ranking
To evaluate the performance of our first year CS majored students, we consider their grades of three ...
- PTA (Advanced Level) 1004 Counting Leaves
Counting Leaves A family hierarchy is usually presented by a pedigree tree. Your job is to count tho ...
- PTA (Advanced Level) 1020 Tree Traversals
Tree Traversals Suppose that all the keys in a binary tree are distinct positive integers. Given the ...
- PTA(Advanced Level)1075.PAT Judge
The ranklist of PAT is generated from the status list, which shows the scores of the submissions. Th ...
- PTA (Advanced Level) 1009 Product of Polynomials
1009 Product of Polynomials This time, you are supposed to find A×B where A and B are two polynomial ...
- PTA (Advanced Level) 1008 Elevator
Elevator The highest building in our city has only one elevator. A request list is made up with Npos ...
- PTA (Advanced Level) 1007 Maximum Subsequence Sum
Maximum Subsequence Sum Given a sequence of K integers { N1, N2, ..., NK }. A continuous su ...
随机推荐
- 教你如何学python
首先,你要有自信心,要明确学习目的.学Python,可以解决在软件使用中所遇到的问题,可以为找到理想工作添加重要砝码.还能锻炼思维,使我们的逻辑思维更加严密:能不断享受到创新的乐趣,将走在高科技的前沿 ...
- git@oschina使用入门(图形界面版)
首先,如果你想使用git@oschina ,你的电脑上必须先有git工具:你可以从这里获取谷歌提供的git.exe http://git-scm.com/当然,如果你能熟练通过命令行操作git,那么这 ...
- Android实现表单提交,webapi接收
1.服务端采用的是.net的WEBAPI接口. 2.android多文件上传. 以下为核心代码: package com.example.my.androidupload; import androi ...
- LightOJ 1213 Fantasy of a Summation(规律 + 快数幂)
http://lightoj.com/volume_showproblem.php?problem=1213 Fantasy of a Summation Time Limit:2000MS ...
- 深入了解java虚拟机(JVM) 第八章 常见的jvm调优策略
一般来说,jvm的调优策略是没有一种固定的方法,只有依靠我们的知识和经验来对项目中出现的问题进行分析,正如吉德林法则那样当你已经把问题清楚写出来,就已经解决了一半.虽然JVM调优中没有固定的策略,但是 ...
- [agc006f] Blackout 神题
Description 给你一个NN行NN列的网格,第ii行第jj列的格子用(i,j)(i,j)表示 一开始的时候有MM个格子被涂成黑色,其他的格子都是白色,具体一点,涂成黑色的格子为(a1,b1 ...
- [USACO17FEB]Why Did the Cow Cross the Road III P(CDQ分治)
题意 两列$n$的排列,相同的数连边,如果一对数有交叉且差的绝对值$>k$,则$++ans$,求$ans$ 题解 可以把每一个数字看成一个三元组$(x,y,z)$,其中$x$表示在第一列的位置, ...
- Docker的安装与启动教程
一.安装Docker Docker官方建议在Ubuntu中安装,因为Docker是基于Ubuntu发布的,而且一般Docker出现的问题Ubuntu是最先更新或者打补丁的.在很多版本的CentOS中是 ...
- 表格Table宽度设置无效的解决方法
表格Table宽度设置无效的解决方法 bootstrap中使用table时发现不管用width赋值方式都无法改变table>td的宽度 解决方法: 设置table:table-layout:fi ...
- dbus-launch(转)
*NAME* dbus-launch - Utility to start a message bus from a shell script dbus-launch - 从shell脚本启动一个消息 ...