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1083.ListGrades(25)水题 PAT(AdvancedLevel)Practise

作者:  发布日期:2016-03-21 21:17:53
  • 题目信息

    1083. List Grades (25)

    时间限制400 ms
    内存限制65536 kB
    代码长度限制16000 B
    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

    解题思路

    排序

    AC代码

    #include <iostream>
    #include <map>
    using namespace std;
    int main()
    {
        ios::sync_with_stdio(false);
        cin.tie(0);
        int n, a, L, R;
        string s1, s2;
        map<int, pair<string, string> > mp;
        cin >> n;
        while(n--){
            cin >> s1 >> s2 >> a;
            mp[a] = make_pair(s1, s2);
        }
        cin >> L >> R;
        bool flag = false;
        for (map<int, pair<string, string> >::reverse_iterator it = mp.rbegin(); it != mp.rend(); ++it){
            if (it->first >= L && it->first <= R){
                cout <<it->second.first <<' ' <<it->second.second <<endl;
                flag = true;
            }
        }
        if (!flag) cout <<'NONE' <<endl;
        return 0;
    }
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