【PAT甲级】1153 Decode Registration Card of PAT
【PAT题解集合】
1153 Decode Registration Card of PAT
题目详情 - 1153 Decode Registration Card of PAT (pintia.cn)
中文翻译
A registration card number of PAT consists of 4 parts:
- the 1st letter represents the test level, namely,
T
for the top level,A
for advance andB
for basic;- the 2nd - 4th digits are the test site number, ranged from 101 to 999;
- the 5th - 10th digits give the test date, in the form of
yymmdd
;- finally the 11th - 13th digits are the testee’s number, ranged from 000 to 999.
Now given a set of registration card numbers and the scores of the card owners, you are supposed to output the various statistics according to the given queries.
Input Specification:
Each input file contains one test case. For each case, the first line gives two positive integers N (≤104) and M (≤100), the numbers of cards and the queries, respectively.
Then N lines follow, each gives a card number and the owner’s score (integer in [0,100]), separated by a space.
After the info of testees, there are M lines, each gives a query in the format
Type Term
, where
Type
being 1 means to output all the testees on a given level, in non-increasing order of their scores. The correspondingTerm
will be the letter which specifies the level;Type
being 2 means to output the total number of testees together with their total scores in a given site. The correspondingTerm
will then be the site number;Type
being 3 means to output the total number of testees of every site for a given test date. The correspondingTerm
will then be the date, given in the same format as in the registration card.Output Specification:
For each query, first print in a line
Case #: input
, where#
is the index of the query case, starting from 1; andinput
is a copy of the corresponding input query. Then output as requested:
- for a type 1 query, the output format is the same as in input, that is,
CardNumber Score
. If there is a tie of the scores, output in increasing alphabetical order of their card numbers (uniqueness of the card numbers is guaranteed);- for a type 2 query, output in the format
Nt Ns
whereNt
is the total number of testees andNs
is their total score;- for a type 3 query, output in the format
Site Nt
whereSite
is the site number andNt
is the total number of testees atSite
. The output must be in non-increasing order ofNt
’s, or in increasing order of site numbers if there is a tie ofNt
.If the result of a query is empty, simply print
NA
.Sample Input:
8 4 B123180908127 99 B102180908003 86 A112180318002 98 T107150310127 62 A107180908108 100 T123180908010 78 B112160918035 88 A107180908021 98 1 A 2 107 3 180908 2 999
Sample Output:
Case 1: 1 A A107180908108 100 A107180908021 98 A112180318002 98 Case 2: 2 107 3 260 Case 3: 3 180908 107 2 123 2 102 1 Case 4: 2 999 NA
思路
PAT
准考证号由 4
部分组成:
- 第
1
位是级别,即T
代表顶级;A
代表甲级;B
代表乙级; - 第
2∼4
位是考场编号,范围从101
到999
; - 第
5∼10
位是考试日期,格式为年、月、日顺次各占2
位; - 最后
11∼13
位是考生编号,范围从000
到999
。
具体思路如下:
- 输入所有的准考证信息,用一个结构体数组存储,方便后续排序使用。
- 每个样例都先输出
Case T: t c
。- 如果
type==1
,需要我们按照分数降序输出,如果分数相等则按照id
的字典序从小到大输出。 - 如果
type==2
,需要输出与考场编号c
相同的所有人的成绩总和。 - 如果
type==3
,需要输出与考试日期c
相同的考场编号以及对应人数。
- 如果
- 注意,如果上面需要输出的信息不存在,则输出
NA
即可。
代码
#include<bits/stdc++.h>
using namespace std;
const int N = 10010;
int n, m;
struct Person
{
string id;
int grade;
bool operator <(const Person& p)const
{
if (grade != p.grade) return grade > p.grade; //按照分数降序
return id < p.id; //按照id字典序升序
}
}p[N];
int main()
{
cin >> n >> m;
//输入所有准考证信息
for (int i = 0; i < n; i++) cin >> p[i].id >> p[i].grade;
for (int i = 1; i <= m; i++)
{
string t, c;
cin >> t >> c;
printf("Case %d: %s %s\n", i, t.c_str(), c.c_str());
if (t == "1")
{
vector<Person> persons;
for (int i = 0; i < n; i++) //将符合等级的加入数组中
if (p[i].id[0] == c[0])
persons.push_back(p[i]);
sort(persons.begin(), persons.end()); //排序
if (persons.empty()) puts("NA");
else
for (auto x : persons)
printf("%s %d\n", x.id.c_str(), x.grade);
}
else if (t == "2")
{
int sum = 0, cnt = 0;
for (int i = 0; i < n; i++)
if (p[i].id.substr(1, 3) == c)
sum += p[i].grade, cnt++;
if (!cnt) puts("NA");
else printf("%d %d\n", cnt, sum);
}
else
{
//先用哈希表对考场编号进行分类
unordered_map<string, int> hash;
for (int i = 0; i < n; i++)
if (p[i].id.substr(4, 6) == c)
hash[p[i].id.substr(1, 3)]++;
//再用容器进行排序操作
vector<pair<int, string>> rooms;
for (auto x : hash) //加负号相当于进行降序操作
rooms.push_back({ -x.second,x.first });
sort(rooms.begin(), rooms.end());
if (rooms.empty()) puts("NA");
else
for (auto x : rooms)
printf("%s %d\n", x.second.c_str(), -x.first);
}
}
return 0;
}