PAT甲级——A1036 Boys vs Girls
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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??−grade?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:
3
Joe M Math990112 89
Mike M CS991301 100
Mary F EE990830 95
Sample Output 1:
Mary EE990830
Joe Math990112
6
Sample Input 2:
1
Jean M AA980920 60
Sample Output 2:
Absent Jean AA980920 NA
1 #include <iostream> 2 #include <vector> 3 #include <algorithm> 4 #include <string> 5 using namespace std; 6 int N; 7 struct Node 8 9 string name, gender, ID; 10 int grade; 11 node; 12 int main() 13 14 cin >> N; 15 vector<Node>male, female; 16 17 //此代码是将数据保存下来排序,还可以直接在输入时就得到最高分和最低分,这样就大大节省了空间和时间 18 for (int i = 0; i < N; ++i) 19 20 cin >> node.name >> node.gender >> node.ID >> node.grade; 21 if (node.gender == "M") 22 male.push_back(node); 23 else 24 female.push_back(node); 25 26 sort(male.begin(), male.end(), [](Node a, Node b) return a.grade < b.grade; ); 27 sort(female.begin(), female.end(), [](Node a, Node b) return a.grade > b.grade; ); 28 if (female.size() == 0) 29 cout << "Absent" << endl; 30 else 31 cout << female[0].name << " " << female[0].ID << endl; 32 if (male.size() == 0) 33 cout << "Absent" << endl; 34 else 35 cout << male[0].name << " " << male[0].ID << endl; 36 if (female.size() == 0 || male.size() == 0) 37 cout << "NA" << endl; 38 else 39 cout << female[0].grade - male[0].grade << endl; 40 return 0; 41
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