• 1083. List Grades (25)


    题目如下:

    Given a list of N student records with name, ID and grade. You are supposed to sort the records with respect to the grade in non-increasing order, and output those student records of which the grades are in a given interval.

    Input Specification:

    Each input file contains one test case. Each case is given in the following format:

    N
    name[1] ID[1] grade[1]
    name[2] ID[2] grade[2]
    ... ...
    name[N] ID[N] grade[N]
    grade1 grade2
    

    where name[i] and ID[i] are strings of no more than 10 characters with no space, grade[i] is an integer in [0, 100], grade1 and grade2 are the boundaries of the grade's interval. It is guaranteed that all the grades are distinct.

    Output Specification:

    For each test case you should output the student records of which the grades are in the given interval [grade1, grade2] and are in non-increasing order. Each student record occupies a line with the student's name and ID, separated by one space. If there is no student's grade in that interval, output "NONE" instead.

    Sample Input 1:
    4
    Tom CS000001 59
    Joe Math990112 89
    Mike CS991301 100
    Mary EE990830 95
    60 100
    
    Sample Output 1:
    Mike CS991301
    Mary EE990830
    Joe Math990112
    
    Sample Input 2:
    2
    Jean AA980920 60
    Ann CS01 80
    90 95
    
    Sample Output 2:
    NONE
    


    简单的排序题,将输入的数据按照成绩降序排序,然后过滤着输出,没有输出输出NONE。

    #include <iostream>
    #include <vector>
    #include <algorithm>
    #include <stdio.h>
    
    using namespace std;
    
    struct Person{
    
        string name,id;
        int grade;
    
        bool operator < (const Person& p) const{
            return grade > p.grade;
        }
    
        Person(string _n, string _id, int _g){
            name = _n;
            id = _id;
            grade = _g;
        }
    
    };
    
    int main()
    {
        int N;
        string name,id;
        int grade;
        vector<Person> persons;
        cin >> N;
        for(int i = 0; i < N; i++){
            cin >> name >> id >> grade;
            persons.push_back(Person(name,id,grade));
        }
        int grade1,grade2;
        cin >> grade1 >> grade2;
        sort(persons.begin(),persons.end());
        bool hasData = false;
        for(int i = 0; i < N; i++){
            if(persons[i].grade < grade1 || persons[i].grade > grade2) continue;
            printf("%s %s
    ",persons[i].name.c_str(),persons[i].id.c_str());
            hasData = true;
        }
        if(!hasData){
            cout << "NONE" << endl;
        }
        return 0;
    }
    


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  • 原文地址:https://www.cnblogs.com/aiwz/p/6154062.html
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