PAT甲级-1153-卡时间模拟题

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

200ms的限时，前后大改了好几次。

Notes：string直接比较优于string.compare()；string转int用stoi()优于通过stringstream转化。

#include <stdio.h>
#include <algorithm>
#include <stdlib.h>
#include <malloc.h>
#include <string.h>
#include <iostream>
#include <queue>
#include <string>
#include <vector>
#include <map>
#include <stack>
#include <sstream>
using namespace std;
const int maxn = 10050;
const int INF = 0x3f3f3f3f;

int n, m;
string str;
int score;
int sum, num, cnt;
int acount[maxn];
map<string, int>mp;

struct node {
	string id, type, site, date;
	int score;
}stu[maxn];

struct node_2 {
	int site;
	int num;
}site;
vector<node_2> vec;

void Add(string str, int score, int i) {
	stu[i].id = str;
	stu[i].score = score;
	stu[i].type = str.substr(0, 1);
	stu[i].site = str.substr(1, 3);
	stu[i].date = str.substr(4, 6);
}

bool cmp(node a, node b) {
	if (a.score == b.score) return a.id < b.id;
	return a.score > b.score;
}

bool cmp2(node_2 a, node_2 b) {
	if (a.num == b.num) return a.site < b.site;
	return a.num > b.num;
}

int main() {
	std::ios::sync_with_stdio(false);
	cin >> n >> m;
	for (int i = 0; i < n; i++) {
		cin >> str >> score;
		Add(str, score, i);
	}
	sort(stu, stu + n, cmp);
	int type;
	string k;
	for (int i = 0; i < m; i++) {
		cin >> type >> k;
		sum = num = cnt = 0;
		cout << "Case " << i + 1 << ": " << type << " " << k << endl;
		if (type == 1) {
			int fg = 0;
			for (int j = 0; j < n; j++) if (stu[j].type == k) { fg = 1; cout << stu[j].id << " " << stu[j].score << endl; }
			if (!fg) cout << "NA" << endl;
		}
		else if (type == 2) {
			for (int j = 0; j < n; j++)
				if (stu[j].site == k) num++, sum += stu[j].score;
			if (!num) { cout << "NA" << endl; continue; }
			cout << num << " " << sum << endl;
		}
		else {
			fill(acount, acount + maxn, 0);
			vec.clear();
			for (int j = 0; j < n; j++) if (stu[j].date == k) acount[stoi(stu[j].site)]++;
			for (int j = 0; j < 1000; j++) {
				if (acount[j]) {
					site.num = acount[j];
					site.site = j;
					vec.push_back(site);
				}
			}
			sort(vec.begin(), vec.end(), cmp2);
			if(vec.empty()) { cout << "NA" << endl; continue; }
			for (int j = 0; j < vec.size(); j++) cout << vec[j].site << " " << vec[j].num << endl;
		}
	}
}