PAT A1153 Decode Registration Card of PAT (25point(s))

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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.

题意

测试样例包含三种查询方法,根据查询给出相应的结果。

解答

查询1:遍历所有学生,找出符合条件(A、B、C)的学生。
查询2:遍历所有学生,找出符合条件的学生,计算总人数和总分数。
查询3:遍历所有学生,如果学生符合条件,计算器加一,这里用unordered_map存储每个考场的学生人数,如果使用map会超时,因为map存储的键是有序的。

#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;
}