PAT_A1153#Decode Registration Card of PAT
Source:
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 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; andinput
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
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 }