PAT(Advanced)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 $M$ ($\le 100$), 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

题目解读：（太长了我就直接扒翻译了）

每个输入文件包含一个测试用例。 对于每种情况，第一行给出两个正整数N和M分别是卡数和查询数。然后N行，每行给出一个卡号和所有者得分（[0,100]中的整数），中间用空格隔开。在被测者的信息之后，有M行，每行以Type Term格式提供查询，其中Type为1表示以给定水平上的所有受测者，以其分数的从小到大的顺序输出。 相应的术语将是指定级别的字母；类型为2表示在给定站点中输出受测者的总数及其总分。 相应的条款将成为站点编号；Type为3表示输出给定测试日期每个站点的受测者总数。 相应的条款将是日期，以与注册卡中相同的格式给出。

解题思路：是乙级的最后两道题，所以基本没怎么考复杂的算法和结构，就是排序。这道题卡在输出上很久，自己做的bug的地方有这几个:

1.运行超时，修改了printf和scanf后还超时，把map换成了unordered_map就不超了（记一下英文库名）

2.答案错误，因为读入日期和考场号是int型，所以还得转成3位和6位输出，高位补0，用%0xd。（所以我为啥不用字符串读呢？？）

3.之前还有一个小bug，scanf("%c",&ch)的时候会读ch为空格，这个时候只需要在前面再补个空格，比如scanf(" %c",&ch)就可以了。（所以我为啥不用字符串读呢！！！）

4.对于map的排序方法——可以把map的东西扔进vector再排序……另外结构体其实不需要构造函数这么麻烦的东西，vector.push_back({string,int}); 这种它会自己帮你组成这个结构体。（大  好  评）

感觉这道题有很多可以化简优化的地方，看了别人的代码也受益匪浅。总之没啥难度，但是坑不少，如果做的越复杂就越容易出错。

AC代码

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <unordered_map>
using namespace std;
struct node {
	string name;
	int score;
};
bool cmp(const node &a,const node &b) {
	if (a.score == b.score)
		return a.name < b.name;
	return a.score > b.score;
}

int main()
{
	int n, m;
	int sum = 0;
	int count = 0;
	vector<node> stu;
	scanf("%d %d", &n, &m);
	for (int i = 0; i < n; i++) {
		char name[100];
		int sc;
		scanf("%s %d", name,&sc);
		stu.push_back({ name, sc });
	}
	sort(stu.begin(), stu.end(), cmp);
	for (int time = 1; time <= m; time++) {
		int type;
		int flag = 0;
		scanf("%d", &type);
		printf("Case %d: %d ", time,type);
		switch (type) {
		case 1:
			char ch;
			scanf(" %c" ,&ch);
			printf("%c\n", ch);
			for (int i = 0; i < n; i++) {
				if (stu[i].name[0] == ch) {
					printf("%s %d\n", stu[i].name.c_str(), stu[i].score);
					flag = 1;
				}
			}
			break;
		case 2:
			int num;
			sum = 0, count = 0;
			scanf(" %d", &num);
			printf("%03d\n", num);
			for (int i = 0; i < n; i++) {
				if (stoi(stu[i].name.substr(1, 3)) == num) {
					count++;
					sum += stu[i].score;
					flag = 1;
				}
			}
			if (flag)
				printf("%d %d\n", count, sum);
			break;
		case 3:
			int date;
			scanf(" %d", &date);
			printf("%06d\n", date);
			unordered_map<string, int> temp;
			for (int i = 0; i < n; i++) {
				if ( stoi(stu[i].name.substr(4, 6)) == date ) {
					temp[stu[i].name.substr(1,3)] ++;
					flag = 1;
				}
			}
			vector<node> tempv;
			for (auto& value : temp)
				tempv.push_back({ value.first, value.second });
			sort(tempv.begin(), tempv.end(), cmp);
			for (auto& value : tempv)
				printf("%s %d\n", value.name.c_str(),value.score);
			break;
		}
		if (!flag) {
			printf("NA\n");
		}
	}
	return 0;
}