PAT日志 1036

最新推荐文章于 2021-07-16 07:28:39 发布

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最新推荐文章于 2021-07-16 07:28:39 发布

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分类专栏： PAT日志文章标签： PAT 1036

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顽强的小白

1036 Boys vs Girls （25 分）

is 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.

Specification:

ach 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.

out pecification:

or 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 gradeF−gradeM. 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

题目解析

题意要求求女生的最高分，男生的最低分，输出两者的差
题目实际考察的是结构体的应用

注意点：姓名的数组名必须超过10位,不然就会过不去最后两个测试点。
数组开大一点没关系。

代码实现

#include <cstdio>
#include <cstring>
 using namespace std;
 
 struct student{
  char name[15];
  char gender;
  char id[20];
  int grade; 
 };
 
 int main(){
  int n,maxF=-1,minM=101,fID,mID;
  student stu[1000];
  scanf("%d",&n);
  for(int i=0;i<n;++i){
   scanf("%s %c %s %d",stu[i].name,&stu[i].gender,stu[i].id ,&stu[i].grade );
   if(stu[i].gender=='M'){
    if(stu[i].grade<minM){
     minM=stu[i].grade;
     mID=i;
    }
   }else{
    if(stu[i].grade>maxF){
     maxF=stu[i].grade;
     fID=i;
    }
   }
  }
  if(maxF==-1) printf("Absent\n");
  else printf("%s %s\n",stu[fID].name,stu[fID].id);
  if(minM==101) printf("Absent\n");
  else printf("%s %s\n",stu[mID].name,stu[mID].id);
  if(maxF==-1||minM==101) printf("NA\n");
  else printf("%d\n",stu[fID].grade-stu[mID].grade);
  
 }