1153 Decode Registration Card of PAT (25 point(s))

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最新推荐文章于 2023-05-10 07:45:00 发布

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1153 Decode Registration Card of PAT (25 point(s))

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

排序题。

1. 尽量使用printf和scanf，否则可能超时；

2. map<key,value>结构方便用来遍历计数，但是如果要基于key和value再排序，则需要遍历map，将它pair<key,value>存储成线性结构（如vector）再重载运算符输出。

map<string,int> mp;
vector<pair<string,int> > v = vector(mp.begin(),mp.end());
bool cmp(pair<string,int> p1,pair<string,int> p2){    }

3. 从字符串中读取信息可以用sscanf；

4. 对于命令1，可以在一开始就排好序，在遇到时顺序遍历即可。并且不会影响命令2和3。

Tips：

如果想要在Dev-Cpp里面使用C++11特性的函数，比如刷算法中常用的stoi、to_string、unordered_map、unordered_set、auto这些，需要在设置里面工具-编译选项-编译器-编译时加入“-std=c++11”
原文链接：https://blog.youkuaiyun.com/liuchuo/article/details/82669248

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<cstring>
#include<unordered_map>
using namespace std;
const int MAX = 1e4+7;
int N,M;
struct card{
	string cardNumber;
	string level;int site,date,number;int score;
};
bool cmp(card a,card b){
	if(a.score!=b.score) return a.score>b.score;
	return a.cardNumber<b.cardNumber; 
}
bool cmp2(pair<int,int> p1,pair<int,int> p2) {
    if(p1.second!=p2.second) return p1.second>p2.second;
    else return p1.first<p2.first;
}
card c[MAX];
int main(void){
	cin>>N>>M;
	for(int i=0;i<N;i++){
		cin>>c[i].cardNumber>>c[i].score;
		c[i].level = c[i].cardNumber.substr(0,1);
		c[i].site = stoi(c[i].cardNumber.substr(1,3));
		c[i].date = stoi(c[i].cardNumber.substr(4,6));
		c[i].number = stoi(c[i].cardNumber.substr(10,3));
	}
	int type;string s;
	sort(c,c+N,cmp); 
	for(int i=1;i<=M;i++){
		cin>>type>>s;
		cout<<"Case "<<i<<": "<<type<<" "<<s<<endl; 
		if(1==type){
			bool flag = false; 
			for(int i=0;i<N;i++){
				if(s==c[i].level){
					printf("%s %d\n",c[i].cardNumber.c_str(),c[i].score);
					flag = true;
				} 
			}
			if(!flag) puts("NA");
		}
		else if(2==type){
			int nt=0,ns=0;int key = stoi(s);
			for(int i=0;i<N;i++){
				if(c[i].site==key){
					nt++;
					ns+=c[i].score; 
				}
			}
			if(nt>0) cout<<nt<<" "<<ns<<endl;
			else puts("NA");
		}
		else if(3==type){
			unordered_map<int,int> mp;int key = stoi(s);
			for(int i=0;i<N;i++){
				if(key== c[i].date){
					mp[c[i].site]++;
				}
			} 
			if(mp.size()==0) puts("NA");
			else{
				vector<pair<int,int> >vec(mp.begin(),mp.end());
				sort(vec.begin(),vec.end(),cmp2);
				for(int i=0;i<vec.size();i++){
					printf("%03d %d\n",vec[i].first,vec[i].second);
				}
			}
		} 
	}
	return 0;
}