18冬第二题 PAT甲级 1153 Decode Registration Card of PAT (25分) 鸡贼的题

冷不防 2023-07-05 13:21 3阅读 0赞

# 题目 #

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

# 鸡贼之处 #

1.准考证带字母  
只能用string存准考证，所以还得弄个结构体，存准考证和分数

2.要求排序  
就因为这个排序，一定要用结构体和vector，在自定义一个排序函数

3.收集、计算  
如果是收集（查询1），那只用结构体和vector没有问题。  
如果是计算（查询3），因为结构体和vector的成员值是只读的（我试过了，成员值只能局部改，出了框改不了），只能用map。而map会超时（很惨了），所以只能用unodered\_map了（头文件同名，别忘了加）。

# 代码注释 #

1.substr  
字符串截断，PAT常见操作

2.for(auto i:v)  
如果用不到STL容器的编号就可以用这个c++14的枚举特性

# 满分代码 #

#include<iostream>
    #include<vector>
    #include<unordered_map>
    using namespace std;
    int n,m,k,num=0,total=0;
    string s;
    struct node{
        string id;
        int value;
    };
    bool cmp(node &a,node &b){
        return a.value==b.value?a.id<b.id:a.value>b.value;
    }
    int main(){
        scanf("%d %d",&n,&m);
        vector<node> v(n);
        for(int i=0;i<n;i++){
            cin>>v[i].id>>v[i].value;
        }
        for(int i=1;i<=m;i++){
            cin>>k>>s;
            printf("Case %d: %d %s\n",i,k,s.c_str());
            vector<node> ans;
            if(k==1){
                for(auto it:v){
                    if(it.id.substr(0,1)==s){
                        ans.push_back(it);
                    }
                }
                if(ans.size()==0){
                    printf("NA\n");
                    continue;
                }
                sort(ans.begin(),ans.end(),cmp);
                for(auto j:ans){
                    printf("%s %d\n",j.id.c_str(),j.value);
                }
            }else if(k==2){
                num=0;
                total=0;
                for(auto it:v){
                    if(it.id.substr(1,3)==s){
                        num++;
                        total+=it.value;
                    }
                }
                if(num==0){
                    printf("NA\n");
                    continue;
                }
                printf("%d %d\n",num,total);
            }else if(k==3){
                unordered_map<string, int> ans3;
                for(auto it:v){
                    if(it.id.substr(4,6)==s){
                        ans3[it.id.substr(1,3)]++;
                    }
                }
                if(ans3.size()==0){
                    printf("NA\n");
                    continue;
                }
                for(auto it:ans3){
                    ans.push_back({it.first,it.second});
                }
                sort(ans.begin(),ans.end(),cmp);
                for(auto j:ans){
                    printf("%s %d\n",j.id.c_str(),j.value);
                }
            }
        }
        return 0;
    }

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