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

    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
    
    #include<iostream>
    #include<algorithm>
    using namespace std;
    
    struct student
    {
        string name;
        char sex;
        string id;
        int grade;
    };
    bool cmp(student a,student b)
    {
        return a.grade>b.grade;
    }
    int main()
    {
        int n;
        vector<student> boy;
        vector<student> girl;
        student temp;
        cin>>n;
        for(int i=0;i<n;i++)
        {
            cin>>temp.name>>temp.sex>>temp.id>>temp.grade;
            if(temp.sex == 'F')
                girl.push_back(temp);
            else
                boy.push_back(temp);
           // scanf("%s%s%s%d",&stu[i].name,&stu[i].sex,&stu[i].id,&stu[i].grade);
        }
        //sort(student,student+n);
        sort(boy.begin(),boy.end(),cmp);
        sort(girl.begin(),girl.end(),cmp);
        if(girl.size()==0)
            cout<<"Absent"<<endl;
        else
            cout<<girl[0].name<<' '<<girl[0].id<<endl;
        if(boy.size()==0)
            cout<<"Absent"<<endl;
        else
            cout<<boy[boy.size()-1].name<<' '<<boy[boy.size()-1].id<<endl;
        if(girl.size() == 0 || boy.size() == 0)
            cout<<"NA"<<endl;
        else
            cout<<abs(boy[boy.size() -1].grade - girl[0].grade)<<endl;
        return 0;
    }
  • 相关阅读:
    git使用总结
    [置顶] 生成学习算法、高斯判别分析、朴素贝叶斯、Laplace平滑——斯坦福ML公开课笔记5
    Java调用cmd压缩文件
    Linux下精确控制时间的函数
    ZTESoft 持续集成 编年史 之 持续集成探索---平台选择
    graylog2+syslog-ng+mongodb构建集中管理日志服务器 --转载
    Active Object 并发模式在 Java 中的应用--转载
    分布式计算的模式语言读后感--基线架构
    Spring 中JCA CCI分析--转载
    Getting started with Apache Camel--转载
  • 原文地址:https://www.cnblogs.com/binanry/p/10013455.html
Copyright © 2011-2022 走看看