PAT甲级——A1153 DecodeRegistrationCardofPAT【25】

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 and B 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

Then

After the info of testees, there are

Type being 1 means to output all the testees on a given level, in non-increasing order of their scores. The corresponding Term 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 corresponding Term 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 corresponding Termwill 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 where Nt is the total number of testees and Ns is their total score;
for a type 3 query, output in the format Site Nt where Site is the site number and Nt is the total number of testees at Site. The output must be in non-increasing order of Nt's, or in increasing order of site numbers if there is a tie of Nt.

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

Solution:
　　这道题就是简单的进行分类判断

 1 #include <iostream>
 2 #include <vector>
 3 #include <string>
 4 #include <unordered_map>
 5 #include <algorithm>
 6 using namespace std;
 7 struct node {
 8     string t;
 9     int value;
10 };
11 bool cmp(const node &a, const node &b) {
12     return a.value != b.value ? a.value > b.value : a.t < b.t;
13 }
14 int main() {
15     int n, k, num;
16     string s;
17     cin >> n >> k;
18     vector<node> v(n);
19     for (int i = 0; i < n; i++)
20         cin >> v[i].t >> v[i].value;
21     for (int i = 1; i <= k; i++) {
22         cin >> num >> s;
23         printf("Case %d: %d %s\n", i, num, s.c_str());
24         vector<node> ans;
25         int cnt = 0, sum = 0;
26         if (num == 1) {
27             for (int j = 0; j < n; j++)
28                 if (v[j].t[0] == s[0]) 
29                     ans.push_back(v[j]);
30         }
31         else if (num == 2) {
32             for (int j = 0; j < n; j++) {
33                 if (v[j].t.substr(1, 3) == s) {
34                     cnt++;
35                     sum += v[j].value;
36                 }
37             }
38             if (cnt != 0) 
39                 printf("%d %d\n", cnt, sum);
40         }
41         else if (num == 3) {
42             unordered_map<string, int> m;
43             for (int j = 0; j < n; j++)
44                 if (v[j].t.substr(4, 6) == s) m[v[j].t.substr(1, 3)]++;
45             for (auto it : m) 
46                 ans.push_back({ it.first, it.second });
47         }
48         sort(ans.begin(), ans.end(), cmp);
49         for (int j = 0; j < ans.size(); j++)
50             printf("%s %d\n", ans[j].t.c_str(), ans[j].value);
51         if (((num == 1 || num == 3) && ans.size() == 0) || (num == 2 && cnt ==
52             0)) printf("NA\n");
53     }
54     return 0;
55 }