1153 Decode Registration Card of PAT

1153 Decode Registration Card of PAT （25 分)

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

//1153 Decode Registration Card of PAT （25 分)

#include 
#include 
#include 
#include 
#include 
using namespace std;
struct node {
	string t;
	int val;
};
bool cmp(const node &a, const node &b) {
	return (a.val != b.val ? a.val > b.val:a.t < b.t);
}
int n, m;

int main() {
	cin >> n >> m;

	vectorv(n);
	for (int i = 0; i < n; i++) {
		cin >> v[i].t >> v[i].val;
	}
	for (int i = 1; i <= m; i++) {
		int type;
		int cnt = 0, sum = 0;
		string term;
		cin >> type >> term;
		printf("Case %d: %d %s\n", i, type, term.c_str());
		vectorans;
		if (type == 1) {
			for (int j = 0; j < n; j++)
				if (v[j].t[0] == term[0]) ans.push_back(v[j]);
		}
		else if (type == 2) {
			for (int j = 0; j < n; j++) {
				if (v[j].t.substr(1, 3) == term) {
					cnt++;
					sum += v[j].val;
				}
			}
			if (cnt != 0)
				printf("%d %d\n", cnt, sum);
		}
		else if (type == 3) {
			unordered_mapmap;
			for (int j = 0; j < n; j++) {
				if (v[j].t.substr(4, 6) == term)//since 4 length 6
					map[v[j].t.substr(1, 3)]++;
			}
			for (auto it : map)
				ans.push_back({ it.first,it.second });
		}
		sort(ans.begin(), ans.end(), cmp);
		for (int j = 0; j < ans.size(); j++)
			printf("%s %d\n", ans[j].t.c_str(), ans[j].val);
		if (((type == 1 || type == 3) && ans.size() == 0) || (type == 2 && cnt == 0))
			printf("NA\n");
	}
	return 0;
}