1153 Decode Registration Card of PAT
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; 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
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
#include<stdio.h> #include<algorithm> #include<vector> #include<string> #include<iostream> using namespace std; int fuck[1005]; struct Student { int score; string id; }; struct part { int count=1; int id; }; vector<Student> seq; bool cmp(Student a,Student b) { if(a.score!=b.score) return a.score>b.score; else return a.id<b.id; } bool cmp1(part a,part b) { if(a.count!=b.count) return a.count>b.count; else return a.id<b.id; } int main() { int seqnum; int cxnum; scanf("%d %d",&seqnum,&cxnum); for(int i=0;i<seqnum;i++) { Student temp; cin>>temp.id; cin>>temp.score; seq.push_back(temp); } for(int i=1;i<=cxnum;i++) { vector<Student> result; int mode; string cxid; scanf("%d",&mode); cin>>cxid; if(mode==2) { int sum=0; int count=0; for(int j=0;j<seq.size();j++) { string ttemp; for(int t=1;t<4;t++) { ttemp.push_back(seq[j].id[t]); } if(ttemp==cxid) { sum+=seq[j].score; count++; } } printf("Case %d: %d ",i,mode); cout<<cxid<<endl; if(count!=0) printf("%d %d",count,sum); else printf("NA"); } if(mode==1) { vector<Student> result; for(int j=0;j<seq.size();j++) { if(seq[j].id[0]==cxid[0]) { result.push_back(seq[j]); } } sort(result.begin(),result.end(),cmp); printf("Case %d: %d ",i,mode); cout<<cxid<<endl; if(result.size()!=0) { for(int j=0;j<result.size();j++) { cout<<result[j].id; printf(" %d",result[j].score); if(j!=result.size()-1) printf("\n"); } } else { printf("NA"); } } if(mode==3) { fill(fuck,fuck+1005,0); bool flag=false; for(int j=0;j<seq.size();j++) { int xx=0; int tt; for(tt=4;tt<10;tt++) { if(cxid[xx++]!=seq[j].id[tt]) break; } if(tt==10) { flag=true; char a[10]; int acount=0; int fucku; for(int mm=1;mm<4;mm++) { a[acount++]=seq[j].id[mm]; } sscanf(a,"%d",&fucku); fuck[fucku]++; } } printf("Case %d: %d ",i,mode); cout<<cxid<<endl; if(flag) { vector<part> aseq; for(int y=0;y<1005;y++) { if(fuck[y]!=0) { part tpart; tpart.count=fuck[y]; tpart.id=y; aseq.push_back(tpart); } } sort(aseq.begin(),aseq.end(),cmp1); for(int t=0;t<aseq.size();t++) { printf("%d %d",aseq[t].id,aseq[t].count); if(t!=aseq.size()-1) printf("\n"); } } else { printf("NA"); } } if(i!=cxnum) printf("\n"); } }