1153 Decode Registration Card of PAT （25 分)

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分类专栏： PAT甲级（二刷）

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解析PAT考试注册卡编号，包含层级、考点、日期和考生号，针对不同查询输出考试统计数据，如按层级列出考生分数、计算特定考点总分或统计指定日期各考点考生数。

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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 (≤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

排序麻烦了点，水题，尽量用scanf printf 否则最后一点超时

#include<stdio.h>
#include<string>
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int n;
struct node{
	string id;
	int score;
}stu[10010];
struct node2{
	int site;
	int num;
};
bool cmp1(node a,node b){
	if(a.score!=b.score){
		return a.score>b.score;
	}
	return a.id<b.id;
}
bool cmp2(node2 a,node2 b){
	if(a.num!=b.num){
		return a.num>b.num;
	}
	return a.site<b.site;
}
void fun1(string term){
	vector<node>a;
	int i;
	for(i=0;i<n;i++){
		if(stu[i].id[0]==term[0]){//同类型放入，然后排序 
			a.push_back(stu[i]);
		}
	}
	sort(a.begin(),a.end(),cmp1);
	if(a.size()){
		for(i=0;i<a.size();i++){
			printf("%s %d\n",a[i].id.c_str(),a[i].score);//c_str()
		} 
	}
	else{
		printf("NA\n");
	}
}
void fun2(string term){
	int i,score=0,num=0;
	for(i=0;i<n;i++){
		if(stu[i].id.substr(1,3)==term){//相同site 
			num++;
			score=score+stu[i].score;
		}
	}
	if(num){
		printf("%d %d\n",num,score);
	}
	else{
		printf("NA\n");
	}
}
void fun3(string term){
	int site[1010]={0},i;
	for(i=0;i<n;i++){
		if(stu[i].id.substr(4,6)==term){
			site[stoi(stu[i].id.substr(1,3))]++;//统计每个site人数 
		}
	}
	vector<node2>a;
	for(i=0;i<1000;i++){
		if(site[i]){
			a.push_back({i,site[i]});//放入site 和  人数 
		}
	}
	sort(a.begin(),a.end(),cmp2);
	if(a.size()){
		for(i=0;i<a.size();i++){
			printf("%d %d\n",a[i].site,a[i].num);
		}
	}
	else{
		printf("NA\n");
	}
}
int main(){
	int m,i,type;
	string term;
	scanf("%d %d",&n,&m);
	for(i=0;i<n;i++){
		cin>>stu[i].id;
		scanf("%d",&stu[i].score);//尽量用scanf，printf,不要用cin,否则最后一点超时 
	}
	for(i=1;i<=m;i++){
		scanf("%d",&type);
		cin>>term;
		printf("Case %d: %d ",i,type);
		cout<<term;
		printf("\n");
		if(type==1){
			fun1(term);
		}
		else if(type==2){
			fun2(term);
		}
		else{
			fun3(term);
		}
	}
}