1036. Boys vs Girls (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

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

Sample Output 1:

Mary EE990830Joe Math9901126

Sample Input 2:

1Jean M AA980920 60

Sample Output 2:

AbsentJean AA980920NA

#include<iostream>#include<string>using namespace std;struct record{string name;char sex;string id;int grade;};int max_of_female(int n,record *re){int i;int which=-1;int max=-1;for(i=0;i<n;i++){if(re[i].sex == 'F')if(which == -1 || re[i].grade>re[which].grade){which = i;max = re[i].grade;}}if(which == -1)cout<<"Absent"<<endl;elsecout<<re[which].name<<' '<<re[which].id<<endl;return max;}int min_of_male(int n,record *re){int i;int which = -1;int min = -1;for(i=0;i<n;i++){if(re[i].sex == 'M')if(which == -1 || re[i].grade<re[which].grade){which = i;min = re[i].grade;}}if(which == -1)cout<<"Absent"<<endl;elsecout<<re[which].name<<' '<<re[which].id<<endl;return min;}int main(){int n,a=-1,b=-1;cin>>n;record* re = new record[n];for(int i=0;i<n;i++)cin>>re[i].name>>re[i].sex>>re[i].id>>re[i].grade;b = max_of_female(n,re);a = min_of_male(n,re);if(a!=-1 && b!=-1)cout<<b-a<<endl;elsecout<<"NA"<<endl;return 0;}