1153 Decode Registration Card of PAT (25分)
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 (≤) and M (≤), the numbers of cards and the queries, respectively.
Then N lines follow, each gives a card number and the owner's score (integer in [), 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; and input
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
解码PAT准考证,我们已知PAT准考证,第一位是TAB(代表基本),二到四是代表考场,五到十位代表日期,最后三位代表考场人数。
我们要进行查询,如果输入1,则查询输入考试级别的分数降序,名字升序排序。如果输入2,则查询该考场人数和总分,输入3,则查询该日期的考场和考场中的人数。
#include <iostream> #include <vector> #include <unordered_map> #include <algorithm> using namespace std; int N, M, choose; string tmp; pair<string, int> cards[11000]; bool cmp(pair<string, int> p1, pair<string, int> p2) { return p1.second != p2.second ? p1.second > p2.second : p1.first < p2.first; } int main() { cin >> N >> M; for(int i = 0; i < N; i++) cin >> cards[i].first >> cards[i].second; for(int c = 1; c <= M; c++) { cin >> choose >> tmp; printf("Case %d: %d %s\n", c, choose, tmp.c_str()); if(choose == 1) { vector<pair<string, int>> v; for(int i = 0; i < N; i++) if(cards[i].first[0] == tmp[0]) v.push_back(cards[i]); if(v.size() != 0) { sort(v.begin(), v.end(), cmp); for(auto x : v) printf("%s %d\n", x.first.c_str(), x.second); } else printf("NA\n"); } else if(choose == 2) { int cnt = 0, all_score = 0; for(int i = 0; i < N; i++) { if(cards[i].first.substr(1, 3) == tmp) { cnt++; all_score += cards[i].second; } } if(cnt != 0) printf("%d %d\n", cnt, all_score); else printf("NA\n"); }else { unordered_map<string, int> m; for(int i = 0; i < N; i++) if(cards[i].first.substr(4, 6) == tmp) m[cards[i].first.substr(1, 3)]++; if(m.size() != 0) { vector<pair<string, int>> v(m.begin(), m.end()); sort(v.begin(), v.end(), cmp); for(auto x: v) printf("%s %d\n", x.first.c_str(), x.second); } else printf("NA\n"); } } return 0; } // B 123 180908 127 // 1 decreasing // 2 Nt Ns // 3 Nnumber testee