1153 Decode Registration Card of PAT -PAT甲级

最新推荐文章于 2023-05-09 15:27:49 发布

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最新推荐文章于 2023-05-09 15:27:49 发布

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本文介绍了一种PAT考试注册卡号的组成方式，包括层级、考场编号、考试日期和考生编号，并详细阐述了如何根据卡号进行成绩统计和查询，包括按层级、考场和日期的不同查询类型。

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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 (≤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

按照题目中的意思分三种查询，第一种查考试的类型，自定义排序顺序；第二种查找考场编号的人数以及总分数，遍历即可；第三种是查考场日期各个考场编号的人数，定义一个结构体进行存储与排序

注意：在每次使用完后须将数据清零，jiegouti，map等都一样，

满分代码如下：

#include<bits/stdc++.h>
using namespace std;
const int N=10005;
map<int,int>mp;
struct Stu{
	string card;
	int tid,tdata,score;//考场编号，考试日期，考生分数 
	Stu(){}
	Stu(string c,int t,int d,int s){
		c=card;
		t=tid;
		tdata=d;
		score=s;
	}
}stu[N];
struct Node{
	int tid;
	int sum=0;
	bool operator < (const Node &a)const{
		if(sum!=a.sum)
			return sum>a.sum;
		return tid<a.tid;
	}
	
}node[1010];
bool cmp1(Stu a,Stu b){
	if(a.score!=b.score)
		return a.score>b.score;
	return a.card<b.card;
}
int n,m;
int main(){
	scanf("%d%d",&n,&m);
	for(int i=0;i<n;i++){
		cin>>stu[i].card>>stu[i].score;
		stu[i].tid=stoi(stu[i].card.substr(1,3));
		stu[i].tdata=stoi(stu[i].card.substr(4,6));
	}
	for(int i=1;i<=m;i++){
		int x;
		scanf("%d",&x);
		if(x==1){
			char c;
			cin>>c;
			vector<Stu>ve;
			for(int j=0;j<n;j++){
				if(stu[j].card[0]==c){
					ve.push_back(stu[j]);
				}
			}
			printf("Case %d: %d %c\n",i,x,c);
			if(ve.empty()){
				printf("NA\n");
			}else{
				sort(ve.begin(),ve.end(),cmp1);
				for(int j=0;j<ve.size();j++){
					printf("%s %d\n",ve[j].card.c_str(),ve[j].score);
				}
			}	
		}else if(x==2){
			int tid;//考场编号
			scanf("%d",&tid);
			int sum=0,num=0;
			for(int j=0;j<n;j++){
				if(stu[j].tid==tid){
					num++;
					sum+=stu[j].score;
				}
			}
			printf("Case %d: %d %03d\n",i,x,tid);
			if(num==0){
				printf("NA\n");
			}else{
				printf("%d %d\n",num,sum);
			}
		}else{
			mp.clear();
			int num=0;
			int tdata;
			scanf("%d",&tdata);
			for(int j=0;j<n;j++){
				if(stu[j].tdata==tdata){
					if(mp[stu[j].tid]==0){
						num++;
						mp[stu[j].tid]=num;
						node[mp[stu[j].tid]].tid=stu[j].tid;
					}
					node[mp[stu[j].tid]].sum++;	
				}
			}
			printf("Case %d: %d %06d\n",i,x,tdata);
			if(num==0){
				printf("NA\n");
			}else{
				sort(node+1,node+num+1);
				for(int j=1;j<=num;j++){
					printf("%d %d\n",node[j].tid,node[j].sum);
				}
				for(int j=1;j<=num;j++){
					node[j].sum=0;
					node[j].tid=0;
				}
			}	
		}
	}
	return 0;
}