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.

## 解答

#include<iostream>
#include<vector>
#include<string>
#include<algorithm>
#include<unordered_map>

using namespace std;

struct student {
string id;
int score;
};

bool cmp1(student s1, student s2)
{
if (s1.score == s2.score) return s1.id < s2.id;
return s1.score > s2.score;
}

int main()
{
int cardNum, queryNum, queryType;
string inputStr;
cin >> cardNum >> queryNum;
vector<student> students(cardNum);

for (int i = 0; i < cardNum; i++)
cin >> students[i].id >> students[i].score;

for (int i = 0; i < queryNum; i++) {
cin >> queryType >> inputStr;
printf("Case %d: %d %s\n", i + 1, queryType, inputStr.c_str());
vector<student> ans;
int totalScore = 0, cnt = 0;
if (queryType == 1) {
for (int i = 0; i < cardNum; i++)
if (students[i].id.substr(0, 1) == inputStr)
ans.push_back(students[i]);
}
else if (queryType == 2) {
for (int i = 0; i < cardNum; i++)
if (students[i].id.substr(1, 3) == inputStr) {
totalScore += students[i].score;
cnt++;
}
}
else {
unordered_map<string, int> mp;
for (int j = 0; j < cardNum; j++)
if (students[j].id.substr(4, 6) == inputStr) mp[students[j].id.substr(1, 3)]++;
for (auto it : mp) ans.push_back({ it.first, it.second });
}
if ((ans.size() == 0 && (queryType == 1 || queryType == 3)) ||
(queryType == 2 && cnt == 0))
printf("NA\n");
else if (queryType == 2)
printf("%d %d\n", cnt, totalScore);
else {
sort(ans.begin(), ans.end(), cmp1);
for (int j = 0; j < ans.size(); j++)
printf("%s %d\n", ans[j].id.c_str(), ans[j].score);
}
}
return 0;
}