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 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 (≤104) 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

浅析

重点关注第三类

Code

#include 
#include 
#include 
#include 
#include 
using namespace std;

struct student{
    string id;
    int score;
};
bool cmp(student a,student b){
    if(a.score!=b.score)
        return a.score>b.score;
    else
        return a.id < b.id;
}
int main(){
    int n, m;
    cin >> n >> m;
    vector v(n);
    for (int i = 0; i < n;i++){
        cin >> v[i].id >> v[i].score;
    }
    for (int i = 1; i <= m;i++){< span>
        int num,cnt=0,sum=0;
        string s;
        cin >> num >> s;
        printf("Case %d: %d %s\n", i, num, s.c_str());
        vector temp;
        if(num==1){
            for (int j = 0; j < n;j++)
                if(v[j].id[0]==s[0])
                    temp.push_back(v[j]);
        }else if(num==2){
            for(int j=0; j < n;j++)
                if(s==v[j].id.substr(1,3)){
                    cnt++;
                    sum+=v[j].score;
                }
            if(cnt!=0){
                printf("%d %d\n", cnt, sum);
                continue;
            }
        }else if(num==3){
            //重点看这里，柳神🐂🍺！！！
            unordered_map<string, int> m;
            for (int j = 0; j < n; j++)
                if (v[j].id.substr(4, 6) == s) 
                    m[v[j].id.substr(1, 3)]++;
            for (auto it : m)
                temp.push_back({it.first, it.second});
        }
        sort(temp.begin(), temp.end(), cmp);
        if (((num == 1 || num == 3) && temp.size() == 0) || (num == 2 && cnt == 0)){
            printf("NA\n");
            continue;
        }
        for(int j = 0; j < temp.size(); j++){
            printf("%s %d\n", temp[j].id.c_str(), temp[j].score);
        }
    }
    return 0;
}