1083. List Grades (25)【水题】——PAT (Advanced Level) Practise
来源:互联网 发布:国云数据裁员 编辑:程序博客网 时间:2024/05/29 02:45
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;}
- 1083. List Grades (25)【水题】——PAT (Advanced Level) Practise
- PAT (Advanced Level) Practise 1083 List Grades (25)
- PAT (Advanced Level) Practise 1083 List Grades (25)
- 【PAT (Advanced Level)】1083. List Grades (25)
- 【PAT】【Advanced Level】1083. List Grades (25)
- PAT (Advanced Level) 1083. List Grades (25) 结构体排序
- PAT (Advanced) 1083. List Grades (25)
- Pat(Advanced Level)Practice--1083(List Grades)
- 1028. List Sorting (25)【排序】——PAT (Advanced Level) Practise
- 1039. Course List for Student (25)【排序】——PAT (Advanced Level) Practise
- 1047. Student List for Course (25)【排序】——PAT (Advanced Level) Practise
- 1052. Linked List Sorting (25)【链表+排序】——PAT (Advanced Level) Practise
- 1074. Reversing Linked List (25)【链表翻转】——PAT (Advanced Level) Practise
- 1097. Deduplication on a Linked List (25)【链表】——PAT (Advanced Level) Practise
- 1003. Emergency (25)——PAT (Advanced Level) Practise
- 1010. Radix (25)——PAT (Advanced Level) Practise
- 1016. Phone Bills (25)——PAT (Advanced Level) Practise
- 1020. Tree Traversals (25)——PAT (Advanced Level) Practise
- 最小生成树kruskal
- mongodb调优那些事(一)-系统设置
- Spinner报错: java.lang.RuntimeException: setOnItemClickListener cannot be used with a spinner.
- java 找出n个元素数组中重复次数最多的数(假设出现次数大于n/2)
- 1082. Read Number in Chinese (25)【字符串处理】——PAT (Advanced Level) Practise
- 1083. List Grades (25)【水题】——PAT (Advanced Level) Practise
- linux epoll 简介
- 易互娱2017实习生招聘在线笔试第一场题目1 : 电子数字
- 理解缓存
- 分享一个连接
- 1084. Broken Keyboard (20)【字符串操作】——PAT (Advanced Level) Practise
- vs2015+DDK7.0开发Win7 64位驱动
- 了解StringBuffer的基本使用方法
- java.lang.AbstractMethodError:oracle.jdbc.T4CConnection.createClob()Ljava