pat 1153

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 (≤) and M (≤), the numbers of cards and the queries, respectively.

Then N lines follow, each gives a card number and the owner's score (integer in [), 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

题意：给定n个pat考生的考号，根据给定的关键字统计结果

思路：一个数组存放考号，另一个存放成绩，题目要求的排序通过sort函数实现，需要重写cmp函数。

注意点：注意格式，有一个点可能超时，要使用unordered_map，不要使用cout输出

代码如下：

#include<cstdio>
#include<iostream>
#include<unordered_map>
#include<vector>
#include<algorithm> 
using namespace std;
unordered_map<int,int> m;
int n,num;
struct node{
    char card[15];
    int socre;
};
char players[10005][20];
int socre[10005];
bool cmp(int& a,int& b){
    if(socre[a]!=socre[b])
        return socre[a]>socre[b];
    else{
        for(int i=0;i<15;i++){
            if(players[a][i]!=players[b][i])
                return players[a][i]<players[b][i];
                
        }
    }
    return false;
}
bool cmp2(int& a,int& b){
    if(m[a]!=m[b])
        return m[a]>m[b];
    else
        return a<b;
}
void fun(int kth,int type,char term[]){
    printf("Case %d: %d %s
",kth,type,term);
    if(type==1){
        vector<int> v;
        int sum=0;
        char level=term[0];
        for(int i=0;i<n;i++){
            if(players[i][0]==level){
                sum++;
                v.push_back(i);
            }
        }
        if(sum==0)
            printf("NA
");
        else{
            sort(v.begin(),v.end(),cmp);
            for(int i=0;i<v.size();i++){
                printf("%s %d
",players[v[i]],socre[v[i]]);
            }
        }
    }
    else if(type==2){
        int sum=0;
        int total=0;
        char site[5];
        for(int j=0;j<3;j++)
            site[j]=term[j];
        for(int i=0;i<n;i++){
            int mark=0;
            for(int j=1;j<=3;j++){
                if(players[i][j]!=site[j-1]){
                    mark=1;
                    break;
                }
            }
            if(mark==0){
                sum++;
                total+=socre[i];
            }
        }
        if(sum!=0)
            printf("%d %d
",sum,total);
        else    
            printf("NA
");
    }
    else if(type==3){
        int sum=0;
        m.clear();
        vector<int> v;
        char data[10];
        for(int i=4;i<=9;i++){
            data[i]=term[i-4];
        }
        for(int i=0;i<n;i++){
            int mark=0;
            for(int j=4;j<10;j++){
                if(players[i][j]!=data[j]){
                    mark=1;
                    break;
                }
            }
            if(mark==0){
                sum++;
                int site=0;
                for(int j=1;j<=3;j++){
                    site=site*10+players[i][j]-'0';
                }
                if(m.count(site)==0){
                    m[site]=1;
                    v.push_back(site);
                }    
                else{
                    m[site]=m[site]+1;
                }
                    
            }
        }
        if(sum==0)
            printf("NA
");
        else{
            sort(v.begin(),v.end(),cmp2);
            for(vector<int>::iterator it=v.begin();it!=v.end();it++){
                printf("%d %d
",*it,m[*it]);
            }
        }

    }

}
int main(){
    scanf("%d %d",&n,&num);
    for(int i=0;i<n;i++){
        scanf("%s %d",&players[i][0],socre+i);
    }
    int type;
    char term[15];
    for(int i=1;i<=num;i++){
         scanf("%d %s",&type,term);
         fun(i,type,term);
    } 
    return 0;
}