PAT A1036. Boys vs Girls (25)

最新推荐文章于 2025-08-30 21:54:06 发布

原创最新推荐文章于 2025-08-30 21:54:06 发布 · 296 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#查找方式

PAT 专栏收录该内容

125 篇文章

订阅专栏

本程序接收学生信息输入，包括姓名、性别、ID和分数，并输出女性最高分与男性最低分的学生信息及其分数差异。若缺少某性别的学生，则相应输出'Absent'及'NA'。

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

#include <cstdio>
#include <algorithm>
#define Max 123400
using namespace std;
struct student{
 char name[11];
 char gen;
 char Id[11];
 int score;
}S[Max];
int main()
{
	int n,m1=101,m2=-1,f1=0,f2=-1,flag=0,d;
	scanf("%d",&n);
	for(int i=0;i<n;i++)
	{
		scanf("%s %c %s %d",S[i].name,&S[i].gen,S[i].Id,&S[i].score);
		if(S[i].gen=='M')
		{
			if(S[i].score<m1)
			{
				m1=S[i].score;
				m2=i;
			}
		}
		else if(S[i].gen=='F')
		{
			if(S[i].score>f1)
			{
				f1=S[i].score;
				f2=i;
			}
		}
	}
	if(f2==-1)
	{
		flag=1;
		printf("Absent\n");
	}
	else {
		printf("%s %s\n",S[f2].name,S[f2].Id);
	}
	if(m2==-1)
	{
		flag=1;
		printf("Absent\n");
	}
	else {
		printf("%s %s\n",S[m2].name,S[m2].Id);
	}
	if(flag==1) printf("NA\n");
	else {
		d=S[f2].score-S[m2].score;
		printf("%d\n",d);
	}
    system("pause");
	return 0;
}