zoukankan      html  css  js  c++  java
  • 1036. Boys vs Girls (25)

    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 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
    

    代码:
    #include <cstdio>
    #include <cstring>
    #include <iostream>
    #include <algorithm>
    #define Max 105
    using namespace std;
    struct stu
    {
        string name,id;
        char gender;
        int grade;
    }s[10000],a,b;
    int main()
    {
        int n;
        cin>>n;
        a.grade = -1;
        b.grade = 101;
        for(int i = 0;i < n;i ++)
        {
            cin>>s[i].name>>s[i].gender>>s[i].id>>s[i].grade;
            if(s[i].gender == 'F' && s[i].grade > a.grade)a = s[i];
            if(s[i].gender == 'M' && s[i].grade < b.grade)b = s[i];
        }
        if(a.grade != -1)cout<<a.name<<' '<<a.id<<endl;
        else cout<<"Absent"<<endl;
        if(b.grade != 101)cout<<b.name<<' '<<b.id<<endl;
        else cout<<"Absent"<<endl;
        if(a.grade != -1 && b.grade != 101)cout<<a.grade - b.grade<<endl;
        else cout<<"NA"<<endl;
    }
  • 相关阅读:
    Charles网络工具
    查找最长子串的长度(不重复字符)
    KMP算法
    java并发编程实战:第十四章----构建自定义的同步工具
    java并发编程实战:第十二章---并发程序的测试
    java并发编程实战:第二章----线程安全性
    java并发编程实战:第三章----对象的共享
    java并发编程实战:第四章----对象的组合
    java并发编程实战:第五章----基础构建模块
    java并发编程实战:第六章----任务执行
  • 原文地址:https://www.cnblogs.com/8023spz/p/8030050.html
Copyright © 2011-2022 走看看