1153 Decode Registration Card of PAT——PAT甲级

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

 solution:
 

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned int ui;
#define endl '\n'
struct node
{
	string id;
	string type;
	string examno;
	string date;
	string manno;
	int grade;
};
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0),cout.tie(0);
	int n,m;cin>>n>>m;
	vector<node>a(n);
	for(int i=0;i<n;i++)
	{
		cin>>a[i].id>>a[i].grade;
		a[i].type=a[i].id[0];
		a[i].examno=a[i].id.substr(1,3);
		a[i].date=a[i].id.substr(4,6);
		a[i].manno=a[i].id.substr(10,3);
	}
	for(int i=0;i<m;i++)
	{
		int t;string s;
		cin>>t>>s;
		cout<<"Case "<<i+1<<": "<<t<<" "<<s<<endl;
		if(t==1)
		{
			vector<node>ans;
			for(auto i:a)
			{
				if(i.type==s)
				{
					ans.push_back(i);
				}
			}
			if(!ans.size())
			{
				cout<<"NA"<<endl;
			}
			else
			{
				sort(ans.begin(),ans.end(),[](node a,node b)
				{
					if(a.grade==b.grade)return a.id<b.id;
					return a.grade>b.grade;
				});
				for(auto j:ans)
				{
					cout<<j.id<<' '<<j.grade<<endl;
				}
			}
		}
		else if(t==2)
		{
			int sum=0;
			int cnt=0;
			for(auto i:a)
			{
				if(i.examno==s)
				{
					sum+=i.grade;
					cnt++;
				}
			}
			if(!cnt)cout<<"NA"<<endl;
			else
			{
				cout<<cnt<<' '<<sum<<endl;
			}
		}
		else if(t==3)
		{
			unordered_map<string,int>mp;
			for(auto i:a)
			{
				if(i.date==s)
				{
					mp[i.examno]++;
				}
			}
			if(!mp.size())
			{
				cout<<"NA"<<endl;
				continue;
			}
			vector<pair<string,int> >ans(mp.begin(),mp.end());
			sort(ans.begin(),ans.end(),[](pair<string,int>a,pair<string,int>b)
			{
				if(a.second==b.second)return a.first<b.first;
				return a.second>b.second;
			});
			for(auto i:ans)
			{
				cout<<i.first<<' '<<i.second<<endl;
			}
		}
	}
}