题目地址

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 (≤10
​4
​​ ) and M (≤100), the numbers of cards and the queries, respectively.

Then N lines follow, each gives a card number and the owner's score (integer in [0,100]), 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

#include<iostream>              //未AC
#include<algorithm>
#include<string>
#include<math.h>
#include<map>
#include<vector>
using namespace std;
const int maxn=10010;
int n,m,k;
struct node{
    string id;
    int score;
}s[maxn];
struct node1{
    int site, cnt;
}ans[1000];
bool cmp(node &a,node &b){
    if(a.id[0]!=b.id[0]) return a.id[0]<b.id[0];
    else if(a.score!=b.score) return a.score>b.score;
    else return a.id<b.id;
}
bool cmp1(node1 &a,node1 &b){
    if(a.cnt!=b.cnt)return a.cnt>b.cnt;
    else return a.site<b.site;
}
int main(){
    cin>>n>>m;
    for(int i=0;i<n;i++){
        cin>>s[i].id>>s[i].score;
    }
    for(int i=1;i<=m;i++){
          cin>>k;
        printf("Case %d: %d ",i,k);
          if(k==1){
              char q;cin>>q;
              cout<<q<<endl;
            sort(s,s+n,cmp);
            int j=0;
            while(s[j].id[0]!=q&&j<n) j++;          //不加小于0会越界;
            if(j==n){
                cout<<"NA"<<endl;continue;
            }
            for(;j<n&&s[j].id[0]==q;j++) cout<<s[j].id<<" "<<s[j].score<<endl;
          }
        else{
            int q;cin>>q;cout<<q<<endl;
            if(k==2){
                int cnt=0,sco=0;
                for(int j=0;j<n;j++){
                    if(stoi(s[j].id.substr(1,3))==q){
                        cnt++;sco+=s[j].score;
                    }
                }
                if(cnt==0){
                    cout<<"NA"<<endl;continue;
                }
                cout<<cnt<<" "<<sco<<endl;
            }
            else if(k==3){
                int max_s=-1;
                for(int j=0;j<n;j++){
                    if(stoi(s[j].id.substr(4,6))==q) {
                        int t=stoi(s[j].id.substr(1,3));
                        max_s=max(max_s,t);
                        ans[t].site=t;
                        ans[t].cnt++;
                    }
                }
                if(max_s==-1){
                    cout<<"NA"<<endl;continue;
                }
                sort(ans,ans+max_s+1,cmp1);
                for(int j=0;ans[j].cnt!=0;j++){
                    string t=to_string(ans[j].site);
                    while(t.length()<3) t='0'+t;
                    cout<<t<<" "<<ans[j].cnt<<endl;
                }
            }
            else cout<<"NA"<<endl;
        }
    }
    return 0;
}