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  • PTA(Advanced Level)1083.List Grades

    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
    
    思路
    • 用结构体数组存储数据,一次排序之后+一次遍历输出符合条件的人的信息就好了
    代码
    #include<bits/stdc++.h>
    using namespace std;
    struct student
    {
    	string name, id;
    	int grade;
    }a[10010];
    
    bool cmp(student a, student b)
    {
    	return a.grade < b.grade;
    }
    
    int main()
    {
    	int n;
    	cin >> n;
    	for(int i=0;i<n;i++)
    		cin >> a[i].name >> a[i].id >> a[i].grade;
    	sort(a, a+n, cmp);
    	int l, r;
    	bool finded = false;     //表示是否至少有一个输出
    	cin >> l >> r;
    	if(l > r)   swap(l,r);
    	for(int i=n-1;i>=0;i--)     //题目要求的是分数按高到低排,所以倒过来遍历
    	{
    		if(a[i].grade >= l && a[i].grade <= r)
    		{
    			finded = true;
    			cout << a[i].name << " " << a[i].id << endl;
    		}
    	}
    	if(!finded)
    		cout << "NONE";
    	return 0;
    }
    
    
    
    引用

    https://pintia.cn/problem-sets/994805342720868352/problems/994805383929905152

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  • 原文地址:https://www.cnblogs.com/MartinLwx/p/11854791.html
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