• List Grades (25)


    题目描述

    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.

    输入描述:

    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.



    输出描述:

    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.

    输入例子:

    4
    Tom CS000001 59
    Joe Math990112 89
    Mike CS991301 100
    Mary EE990830 95
    60 100

    输出例子:

    Mike CS991301
    Mary EE990830
    Joe Math990112

    思路:简单模拟,n个学生信息包括学生姓名,科目代号,分数,给两个分数的左右区间,问分数在这两个区间的范围内的学生有哪些,有的话按照成绩的非递增顺序输出学生的姓名和科目。
    没有输出“NONE”
    #include<bits/stdc++.h>
    using namespace std;
    
    struct student {
        char name[11];
        int grade;
        char subject[11];
    };
    student res[101];
    student stu[101];
    bool cmp(student a, student b)
    {
        return a.grade >= b.grade;
    }
    int main()
    {
        //freopen("in.txt", "r", stdin);
        ios::sync_with_stdio(false), cin.tie(0);
        int n;
        cin >> n;
        for (int i = 0; i < n; i++)
            cin >> stu[i].name >> stu[i].subject >> stu[i].grade;
        int left, right;
        cin >> left >> right;
        int cnt = 0;
        for (int i = 0; i < n; i++)
            if (stu[i].grade >= left && stu[i].grade <= right)
                res[cnt++] = stu[i];
        if (cnt == 0) {cout << "NONE"; return 0;}
        sort(res, res + cnt, cmp);
        for (int i = 0; i < cnt; i++)
            cout << res[i].name << " " << res[i].subject << endl;
        return 0;
    
    }
    View Code
  • 相关阅读:
    算法14-----位运算操作(1)
    算法13------集合所有的子集或者字符串所有子串
    21、缓存设计
    20、内存溢出(Out of Memory)
    19、内容共享
    14、数据传输更省电
    15、自定义Content Provider
    16、编写适应多个API Level的APK
    17、屏幕适配,多语言支持,手机类型适配
    18、不同平台版本
  • 原文地址:https://www.cnblogs.com/kuroko-ghh/p/10289921.html
Copyright © 2020-2023  润新知