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 91Then 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 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 999999Sample Output
1 C 1 M 1 E 1 A 3 A N/A===============================================
又回来做PAT题目了，C++都忘得差不多干净了有没有！下次尝试下提交Python的答案吧~诶嘿嘿
这道题没什么难的，因为它时间要求不高。
我做了个链表把学生成绩的节点连起来，这样是因为学号有6位数（其实要是弄个1000000行的矩阵也没什么了不起的嘛哼！），反正之后就是链表查找的问题了，没啥可说。
#include <iostream>
using namespace std;
struct RecordNode
{
//存放单个学生数据的链表节点
int studentID;
int results[4]; //0-A, 1-C, 2-M, 3-E
RecordNode* nextNode;
RecordNode(int sid, int c, int m, int e, int a)
{
studentID = sid;
results[0] = a;
results[1] = c;
results[2] = m;
results[3] = e;
nextNode = NULL;
}
};
RecordNode* rootNode = NULL;
RecordNode* getNodePointer(int studentID)
{
//没有则返回NULL
RecordNode* currNode = rootNode;
while (currNode != NULL)
{
if (currNode->studentID == studentID)
return currNode;
currNode = currNode->nextNode;
}
return currNode;
}
void getRankInMatrix(int studentID, int* ranks)
//从矩阵中得到某人的all排位
//(subj: 0-C, 1-M, 2-E, 3-A)
{
//int ranks[4] = { 1, 1, 1, 1 };
RecordNode* node = getNodePointer(studentID);
if (node == NULL)
{
//ranks = NULL;
//delete ranks;
ranks[0] = -1;
return;
}
int* results = node->results; //所有课的成绩
RecordNode* opNode = rootNode;
while (opNode != NULL)
{
//遍历
for (int subj = 0; subj < 4; subj++)
{
//subjects
if (opNode->results[subj] > results[subj])
ranks[subj] ++;
}
opNode = opNode->nextNode;
}
//return ranks;
}
char ACME[] = "ACME";
int main()
{
int M, N;
cin >> N >> M;
RecordNode* opNode = rootNode;
for (int i = 0; i < N; i++)
{
int sid, c, m, e, a;
cin >> sid >> c >> m >> e;
a = (int)(((c + m + e) / 3.0) + 0.5); //四舍五入
if (opNode == NULL)
{
//新建根节点
opNode = new RecordNode(sid, c, m, e, a);
rootNode = opNode;
continue;
}
RecordNode *currNode = new RecordNode(sid, c, m, e, a);
opNode->nextNode = currNode;
opNode = opNode->nextNode; //move forward
}
for (int i = 0; i < M; i++)
{
int sid;
cin >> sid;
int ranks[] = { 1, 1, 1, 1 };
getRankInMatrix(sid, ranks);
if (ranks[0] == -1)
{
cout << "N/A";
if (i < M - 1)
cout << endl;
continue;
}
int bestRank = ranks[0];
int bestRankSubj = 0;
for (int x = 1; x < 4; x++)
{
if (ranks[x] < bestRank)
{
bestRank = ranks[x];
bestRankSubj = x;
}
}
cout << bestRank << " " << ACME[bestRankSubj];
if (i < M - 1)
cout << endl;
}
return 0;
}
