PAT A1012 The Best Rank (25 分)

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 CME 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 (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 CM 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
 
#include <stdio.h>
#include <algorithm>
#include <vector>
using namespace std;
const int maxn = 1000001;
struct stu{
    int id;
    int grade[4];
};
vector<stu> vs;
int myrank[maxn][4];
char list[4] = { 'A', 'C', 'M', 'E' };
int now;
bool cmp(stu s1, stu s2){
    return s1.grade[now] > s2.grade[now];
}
int main(){
    int n, m;
    scanf("%d %d", &n, &m);
    for (int i = 0; i<n; i++){
        int id;
        int c, m, e, a;
        scanf("%d", &id);
        scanf("%d %d %d\n", &c, &m, &e);
        stu s;
        s.id = id;
        s.grade[1] = c;
        s.grade[2] = m;
        s.grade[3] = e;
        a = c + m + e;
        s.grade[0] = a;
        vs.push_back(s);
    }
    for (now = 0; now < 4; now++){
        sort(vs.begin(), vs.end(), cmp);
        myrank[vs[0].id][now] = 1;
        for (int k = 1; k < n; k++){
            if (vs[k].grade[now] != vs[k - 1].grade[now]){
                myrank[vs[k].id][now] = k + 1;
            }
            else{
                myrank[vs[k].id][now] = myrank[vs[k-1].id][now];
            }
        }
    }
    for (int i = 0; i < m; i++){
        int j;
        scanf("%d", &j);
        
        if (myrank[j][0] != 0){
            int best = maxn, best_k = 0;;
            for (int k = 0; k < 4; k++){
                if (myrank[j][k]<best){
                    best_k = k;
                    best = myrank[j][k];
                }
            }
            printf("%d %c\n", myrank[j][best_k], list[best_k]);
        }
        else{
            printf("N/A\n");
        }
    }
}

注意点:很简单的一道题目,却花了一个多小时,一开始题目意思理解错了,虽然就算理解对了,也一样。就觉得很麻烦,要开好多个数组,加好多属性,总觉得可能有捷径,就一直不下手。反思一下,就是要用最笨的最麻烦的方法先写出来,如果超时了,再去想哪里可以改进。这道题就是直接开个记录排名的数组,或者在结构体里加上各个成绩的排名属性,每个都排一下序,记录下来。这里唯一的小技巧是记录成绩用一个数组,而不是4个int,这样写cmp函数只要写一个。

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转载自www.cnblogs.com/tccbj/p/10375160.html
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