PTA甲级——1036

1036 Boys vs Girls

This time you are asked to tell the difference between the lowest grade of all the male students and the highest grade of all the female students.

Input Specification:

Each input file contains one test case. Each case contains a positive integer N, followed by N lines of student information. Each line contains a student’s name, gender, ID and grade, separated by a space, where name and ID are strings of no more than 10 characters with no space, gender is either F (female) or M (male), and grade is an integer between 0 and 100. It is guaranteed that all the grades are distinct.

Output Specification:

For each test case, output in 3 lines. The first line gives the name and ID of the female student with the highest grade, and the second line gives that of the male student with the lowest grade. The third line gives the difference grade F−grad e M. If one such kind of student is missing, output Absent in the corresponding line, and output NA in the third line instead.

Sample Input 1:

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2
3
4
3
Joe M Math990112 89
Mike M CS991301 100
Mary F EE990830 95

Sample Output 1:

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2
3
Mary EE990830
Joe Math990112
6

Sample Input 2:

1
2
1
Jean M AA980920 60

Sample Output 2:

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2
3
Absent
Jean AA980920
NA

思路

​ 大致题意：给你一些男生和女生的成绩，然后需要你先后输出男生的最好成绩同学的姓名和id，女生的最好成绩同学的姓名和id，如果不存在男生或女生就输出Absent。最后还需要输出这两个同学的成绩差值，如果男生或者女生不存在则输出NA即可。

​ 思路：非常明显的思路，就是准备两个node,然后再将数据存放到里面去，最后自定义排序一下就ok了。

代码

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#include <iostream>
#include <algorithm>
#include <cmath>

using namespace std;
const int N = 1e4 + 10;
typedef struct {
    string name;
    string id;
    int grade;
} Node;

Node F[N],M[N];
int k1, k2;

bool cmp1(Node A, Node B) {
    return A.grade > B.grade;
}

bool cmp2(Node A, Node B) {
    return A.grade < B.grade;
}

int main()
{
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++){
        string a, b, c;
        int d;
        cin >> a >> b >> c >> d;
        if (b == "M") {
            M[k2].name = a;
            M[k2].id = c;
            M[k2++].grade = d;
        }
        else {
            F[k1].name = a;
            F[k1].id = c;
            F[k1++].grade = d;
        }
    }

    if (k1) {
        sort(F, F + k1,cmp1);
        cout << F[0].name << " " << F[0].id << '\n';
    } else {
        cout << "Absent\n";
    }

    
    if (k2) {
        sort(M, M + k2,cmp2);
        cout << M[0].name << " " << M[0].id << '\n';
    } else {
        cout << "Absent\n";
    }

    if (k1 && k2){
        cout << F[0].grade - M[0].grade;
    } else cout << "NA";
    return 0;
}