PAT A1036 Boys vs Girls 版本2有个小问题

最新推荐文章于 2020-03-01 11:28:19 发布

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本文介绍了一种算法，用于分析学生数据库中男性和女性学生的最高和最低成绩，并计算两者之间的差距。通过输入学生的姓名、性别、ID和成绩，算法能够找出成绩极端的学生，并给出成绩差异。特别关注了数据输入和结构化处理，确保了所有成绩的独特性。

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题目

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

结构体版本

把MF作为结构题的名称类别

#include <iostream>
#include <algorithm>
#include <string>
using namespace std;
struct person{
    char name[15];
    char id[15];
    int grade;
}M,F,temp;
void init(){
    M.grade=101;
    F.grade=-1;
}
int main(){
    init();
    int n;
    char gender;
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%s %c %s %d",temp.name,&gender,temp.id,&temp.grade);
        if(gender=='M'&&(temp.grade<M.grade))
        {
            M=temp;
        }
        else if((gender=='F'&&(temp.grade>F.grade))){
            F=temp;
        }
    }
    if(F.grade!=-1){
        printf("%s %s\n",F.name,F.id);
    }
    else
        printf("Absent\n");
    
    if(M.grade!=101){
        printf("%s %s\n",M.name,M.id);
    }
    else
        printf("Absent\n");

    if(M.grade!=101&&F.grade!=-1)
        printf("%d\n",abs(M.grade-F.grade));
    else
        printf("NA\n");
        

       
   
    return 0;
}

在这里插入图片描述

scanf 结构体输入的情况下，temp.name. 不是temp.name[I]
直接替换不用分开替换 M=temp；
针对特殊的情况根据初始值判断是否有值后序输入

错误版本

#include <iostream>
#include <algorithm>
#include <string>
using namespace std;
int main(){
    int n;
    char name[15],id[15],NAME[15],ID[15];
    char  gender;
    int temp_score,count,ax;
    int score_M=0,score_F=0;
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%s %c %s %d",&name[i],
              &gender,&id[i],&temp_score);
        if(gender=='M'&&n>1)
        {
            if(temp_score>score_M)
                score_M=temp_score;
            NAME[i]=name[i];
            ax=i;
            ID[i]=id[i];
        }
        else if(gender=='F'&&n>1)
        {
            if(temp_score<score_F)
                score_F=temp_score;
            NAME[i]=name[i];
            ax=i;
            ID[i]=id[i];
        }
    }
        count=abs(score_M-score_F);
        printf("%s %s %d",NAME,ID,count);
        if(n==1){
            
        }
       
   
    return 0;
}