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

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

   题目大意：

         给出n个记录和m次查询。每个记录都有一位的考试类型，3位的课室，6位的年月日，3位考生id。查询包括了两部分，一部分是type， 一部分是内容，如果type == 1， 要求输出该等级的id和分数，按分数从大到小排列，如果分数相同则让id小的排前面。

如果type == 2，就输出指定课室的人数以及总分。如果type == 3 则输出给出的日期，按照人数从大到小，如果人数一直，课室号小的优先。

思路：

      观察他们排序的要求，除了2以外，排序方式都是有一个总分优先，id随后。

      用map<string, int> 来应对第三种情形。

参考代码：

 

 

 

#include<unordered_map>
#include<algorithm>
#include<vector>
#include<iostream>
using namespace std;
struct node{
	string id;
	int v;
};
vector<node>list;
int n, m, flag;
string c;
bool cmp(node a, node b){
	return a.v != b.v ? a.v > b.v : a.id < b.id;
}
int main(){
	scanf("%d%d", &n, &m);
	list.resize(n);
	for(int i = 0; i < n; ++i) cin >> list[i].id >> list[i].v;
	for(int i = 1; i <= m; ++i){
		cin >> flag >> c;
		printf("Case %d: %d %s\n", i , flag, c.c_str());
		vector<node> ans;
		int num = 0, total = 0;
		if(flag == 1){
			for(int j = 0; j < n; ++j)
				if(list[j].id[0] == c[0]) ans.push_back(list[j]);
		} else if(flag == 2){
			for(int j = 0; j < n; ++j)
				if(list[j].id.substr(1, 3) == c) num++, total += list[j].v;
			if(num != 0) printf("%d %d\n", num, total);
		} else {
			unordered_map<string, int > mp;
			for(int j = 0; j < n; ++j){
				string t = list[j].id.substr(4, 6);
				if(t == c) mp[list[j].id.substr(1, 3)]++;
			}
			for(auto it : mp) 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].id.c_str(), ans[j].v);
		if(((flag == 1 || flag == 3) && ans.size() == 0) || (flag == 2 && num == 0))  printf("NA\n");
	}
}

代码参考：https://blog.csdn.net/liuchuo/article/details/84973049