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.

#include
#include
#include
#include
#include
#include
#include 
using namespace std;

class Student{
public:
    string card_id;
    char level;
    string site;
    string date;
    string no;
    int grade;
    Student(string card_id,int grade_){
        this->card_id = card_id;
        level = card_id[0];
        site = card_id.substr(1,3);
        date = card_id.substr(4,6);
        no = card_id.substr(10);
        grade = grade_;
    }
};
const int MAX = 10005;
Student* s[MAX];
vector v;

bool cmp(Student* x, Student* y) {
    if(x->grade!=y->grade) return x->grade>y->grade;//从大到小
    else return x->card_idcard_id;//从小到大
}

bool cmp1(pair p1,pair p2) {
    if(p1.second!=p2.second) return p1.second>p2.second;
    else return p1.first>query;
        printf("Case %d: %d %s\n",i,type,query.c_str());
        if(type==1) {
            for(int i = 0; i < N; i++) {
                if(s[i]->level==query[0]) {
                    v.push_back(s[i]);
                }
            }
            sort(v.begin(),v.end(),cmp);
            if(v.size()==0) printf("NA\n");
            else {
                for(int j = 0; j < v.size(); j++) {
                    printf("%s %d\n",v[j]->card_id.c_str(),v[j]->grade);
                }
            }
        } else if(type==2) {
            int totPerson=0,totScore=0;
            for(int i = 0; i < N; i++) {
                if(s[i]->site==query) {
                    totPerson++;
                    totScore += s[i]->grade;
                }
            }
            if(totPerson!=0)printf("%d %d\n",totPerson,totScore);
            else printf("NA\n");
        } else if(type==3) {
            unordered_map mp;
            for(int i = 0; i < N; i++) {
                if(s[i]->date==query) {
                    mp[s[i]->site]++;
                }
            }
            if(mp.size()==0) printf("NA\n");
            else {
                vector > vec(mp.begin(),mp.end());
                sort(vec.begin(),vec.end(),cmp1);
                for(int i = 0; i< vec.size(); i++) {
                    printf("%s %d\n",vec[i].first.c_str(),vec[i].second);
                }
            }

        }
    }
    return 0;
}

1153 Decode Registration Card of PAT （25 分）

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