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 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 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

 #include <iostream>
 #include <vector>
 #include <string>
 #include <unordered_map>
 #include <algorithm>
 using namespace std;
 struct node {
     string t;
     int value;
 };
 bool cmp(const node &a, const node &b) {
     return a.value != b.value ? a.value > b.value : a.t < b.t;
 }
 int main() {
     int n, k, num;
     string s;
     cin >> n >> k;
     vector<node> v(n);
     for (int i = ; i < n; i++)
         cin >> v[i].t >> v[i].value;
     for (int i = ; i <= k; i++) {
         cin >> num >> s;
         printf("Case %d: %d %s\n", i, num, s.c_str());
         vector<node> ans;
         int cnt = , sum = ;
         if (num == ) {
             for (int j = ; j < n; j++)
                 if (v[j].t[] == s[])
                     ans.push_back(v[j]);
         }
         else if (num == ) {
             for (int j = ; j < n; j++) {
                 if (v[j].t.substr(, ) == s) {
                     cnt++;
                     sum += v[j].value;
                 }
             }
             if (cnt != )
                 printf("%d %d\n", cnt, sum);
         }
         else if (num == ) {
             unordered_map<string, int> m;
             for (int j = ; j < n; j++)
                 if (v[j].t.substr(, ) == s) m[v[j].t.substr(, )]++;
             for (auto it : m)
                 ans.push_back({ it.first, it.second });
         }
         sort(ans.begin(), ans.end(), cmp);
         for (int j = ; j < ans.size(); j++)
             printf("%s %d\n", ans[j].t.c_str(), ans[j].value);
         if (((num ==  || num == ) && ans.size() == ) || (num ==  && cnt ==
             )) printf("NA\n");
     }
     return ;
 }