1083. List Grades (25)【水題】——PAT (Advanced Level) Practise

題目信息

1083. List Grades (25)

時間限制400 ms
內存限制65536 kB
代碼長度限制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

解題思路

排序

AC代碼

#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;
}