PAT_A1153#Decode Registration Card of PAT

Source:

PAT 1153 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

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 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

Code:

 1 /*
 2 Data: 2019-05-12 19:20:45
 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<iostream>
17 #include<map>
18 #include<string>
19 #include<algorithm>
20 using namespace std;
21 const int M=1e4+10;
22 int n,m;
23 struct node
24 {
25     string id;
26     int score;
27 }info[M],ans[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)
38 {
39     int index;
40     string s;
41     cin >> index >> s;
42     printf("Case %d: %d %s\n", k,index,s.c_str());
43     int pos=0,cnt=0,sum=0;
44     if(index == 1)
45     {
46         for(int i=0; i<n; i++)
47             if(info[i].id[0]==s[0])
48                 ans[pos++] = info[i];
49     }
50     else if(index == 2)
51     {
52         for(int i=0; i<n; i++)
53             if(info[i].id.substr(1,3)==s)
54                 {cnt++;  sum += info[i].score;}
55         if(cnt!=0)
56             printf("%d %d\n", cnt,sum);
57     }
58     else if(index == 3)
59     {
60         map<string,int> mp;
61         for(int i=0; i<n; i++)
62             if(info[i].id.substr(4,6)==s)
63                 mp[info[i].id.substr(1,3)]++;
64         for(auto it=mp.begin(); it!=mp.end(); it++)
65             ans[pos++]={it->first,it->second};
66     }
67     if(cnt==0 && pos==0)
68         printf("NA\n");
69     else if(pos!=0)
70     {
71         sort(ans, ans+pos, cmp);
72         for(int i=0; i<pos; i++)
73             printf("%s %d\n", ans[i].id.c_str(), ans[i].score);
74     }
75 }
76 
77 int main()
78 {
79 #ifdef ONLINE_JUDGE
80 #else
81     freopen("Test.txt", "r", stdin);
82 #endif // ONLINE_JUDGE
83 
84     scanf("%d%d", &n,&m);
85     for(int i=0; i<n; i++)
86         cin >> info[i].id >> info[i].score;
87     for(int i=1; i<=m; i++)
88         Query(i);
89 
90     return 0;
91 }

PAT_A1153#Decode Registration Card of PAT

Source:

Description:

Input Specification:

Output Specification:

Sample Input:

Sample Output:

Keys:

Attention:

Code:

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