1153 Decode Registration Card of PAT （25 分）解码pat准考证英文版

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

 可怕，解码pat准考证的中文版本，在我上上一片博客中，有中文版的，也有一些坑的介绍，这里依旧只贴代码

#include<iostream>

#include<string>

#include<vector>

#include <unordered_map>

#include<algorithm>

using namespace std;

struct node{

	string str;

	int score;

};

bool cmp(struct node a,struct node b){

	return a.score != b.score?a.score > b.score:a.str < b.str;

}

int main(){

	int n,m;

	cin >> n >> m;

	vector<node> v(n);

	for(int i = 0;i < n;i++){

		cin >> v[i].str >> v[i].score;

	}

	for(int i = 1;i <= m;i++){

		int k;

		string s;

		cin >> k >> s;

		vector<node> temp;

		printf("Case %d: %d %s\n",i,k,s.c_str());

		if(k == 1){

			for(int j = 0;j < n;j++){

				if(s[0] == v[j].str[0])

					temp.push_back(v[j]);

			}

			sort(temp.begin(),temp.end(),cmp);

			if(temp.size() == 0)

				printf("NA\n");

			else{

				for(int j = 0;j < temp.size();j++){

					printf("%s %d\n",temp[j].str.c_str(),temp[j].score);

				}

			} 

		}else if(k == 2){

			int count = 0,sum = 0;

			for(int j = 0;j < n;j++){

				if(v[j].str.substr(1,3) == s){

					count++;

					sum += v[j].score;

				}

			}

			if(count==0){

				printf("NA\n");

			}else{

				printf("%d %d\n",count,sum);

			}

		}else if(k == 3){

			unordered_map<string,int> m;

			for(int j = 0;j < n;j++){

				if(v[j].str.substr(4,6) == s){

					m[v[j].str.substr(1,3)]++;

				}

			}

			for(auto it:m){

				temp.push_back({it.first,it.second});

			}

			sort(temp.begin(),temp.end(),cmp);

			if(temp.size() == 0)

				printf("NA\n");

			else{

				for(int j = 0;j < temp.size();j++){

					printf("%s %d\n",temp[j].str.c_str(),temp[j].score);

				}

			} 

		}

 

	}

	return 0;

}