PAT_A1153#Decode Registration Card of PAT

Source:

PAT A1153 Decode Registration Card of PAT (25 分)

Description:

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 Term 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 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
NA

Keys:

  • map（C++ STL）
  • string（C++ STL）
  • 快乐模拟

Attention:

  • 每轮mp要记得清空-,-
  • cmp引用传参效率更快
  • cout 输出会超时

Code:

 1 /*
 2 Data: 2019-08-02 21:39:05
 3 Problem: PAT_A1153#Decode Registration Card of PAT
 4 AC: 34:30
 5 
 6 题目大意：
 7 第一行：给出卡片数量n<=1e4，和查询数量m<=100
 8 接下来n行，给出卡号，成绩[0,100]
 9 接下来m行，给出查询
10 1. 给出指定等级，输出各个考生，及其成绩，递减，卡号递增
11 2. 给出指定地点，输出考生总数，及其总分，
12 3. 给出指定时间，输出各个考场的考生总数，人数递减，地点递增
13 4. 查询失败输出NA
14 */
15 #include<cstdio>
16 #include<map>
17 #include<string>
18 #include<vector>
19 #include<iostream>
20 #include<algorithm>
21 using namespace std;
22 const int M=1e4+10;
23 struct node
24 {
25     string id;
26     int score;
27 }info[M];
28 
29 bool cmp(const node &a, const node &b)
30 {
31     if(a.score != b.score)
32         return a.score > b.score;
33     else
34         return a.id < b.id;
35 }
36 
37 void Query(int k, int n)
38 {
39     int index,sum=0,cnt=0;
40     string s;
41     cin >> index >> s;
42     printf("Case %d: %d %s\n", k,index,s.c_str());
43     vector<node> ans;
44     map<string,int> mp;
45     for(int i=0; i<n; i++)
46     {
47         if(index==1 && info[i].id[0]==s[0])
48             ans.push_back(info[i]);
49         else if(index==2 && info[i].id.substr(1,3)==s)
50         {
51             cnt++;
52             sum += info[i].score;
53         }
54         else if(index==3 && info[i].id.substr(4,6)==s)
55             mp[info[i].id.substr(1,3)]++;
56     }
57     if(index==2 && cnt!=0)
58         printf("%d %d\n", cnt, sum);
59     else
60         for(auto it=mp.begin(); it!=mp.end(); it++)
61             ans.push_back({it->first,it->second});
62     sort(ans.begin(),ans.end(),cmp);
63     for(int i=0; i<ans.size(); i++)
64         printf("%s %d\n", ans[i].id.c_str(),ans[i].score);
65     if(ans.size()==0 && cnt==0)
66         printf("NA\n");
67 
68     return;
69 }
70 
71 int main()
72 {
73 #ifdef ONLINE_JUDGE
74 #else
75     freopen("Test.txt", "r", stdin);
76 #endif // ONLINE_JUDGE
77 
78     int n,m;
79     scanf("%d%d", &n,&m);
80     for(int i=0; i<n; i++)
81         cin >> info[i].id >> info[i].score;
82     for(int i=1; i<=m; i++)
83         Query(i,n);
84 
85     return 0;
86 }

 

posted @ 2019-05-12 20:01  林東雨  阅读(401)  评论(0)    收藏  举报