1153 Decode Registration Card of PAT （25 分)

以你之姓@ 2022-03-02 15:56 149阅读 0赞

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

题意：给出n个人，m个查询。n个人的信息有考试种类type，座位号Site，日期data，随机数字num组成，最后是分数。

给出3种询问，1：考试种类为A/B/T的人的排序，先按分数排，再按前面一串的字典序排。

2：座位号为x的人有多少个，其相加总分是多少。

3：日期为x的人座位号分类，按照出现次数多的排序，一样都时候按座位号字典序排。

思路：暴力模拟，用结构体存每个人的信息，输出的时候注意3的时候，日期可能为“020202”，则要%06d输出。

还有1排序要放在while(m--)外面，不然会超时。

#include<stdio.h>
    #include<string.h>
    #include<math.h>
    #include<iostream>
    #include<string>
    #include<sstream>
    #include<algorithm>
    #include<map>
    #include<set>
    #include<queue>
    #include<stack>
    #include<vector>
    using namespace std;
    #define inf 0x3f3f3f3f
    #define LL long long
    struct sb
    {
    	char type;
    	string a;
    	int site;
    	int date;
    	int num;
    	int score;
    }p[10005];
    struct sb3
    {
    	int num,sum;
    }pp[10005];
    bool cmp1(sb u,sb w)
    {
    	//if(u.type<w.type)
    	//return u.type<w.type;
    	if(u.score!=w.score)
    	return u.score>w.score;
    	return u.a<w.a;
    }
    bool cmp3(sb3 u,sb3 w)
    {
    	if(u.sum!=w.sum)
    	return u.sum>w.sum;
    	return u.num<w.num;
    }
    int main()
    {
    	int n,m,i,j,x,y,o=1;
    	char c;
    	scanf("%d%d",&n,&m);
    	for(i=0;i<n;i++)
    	{
    		cin>>p[i].a>>p[i].score;
    		p[i].type=p[i].a[0];
    		p[i].site=(p[i].a[1]-'0')*100+(p[i].a[2]-'0')*10+(p[i].a[3]-'0');
    		p[i].date=(p[i].a[4]-'0')*100000+(p[i].a[5]-'0')*10000+(p[i].a[6]-'0')*1000+(p[i].a[7]-'0')*100+(p[i].a[8]-'0')*10+(p[i].a[9]-'0');
    		p[i].num=(p[i].a[10]-'0')*100+(p[i].a[11]-'0')*10+(p[i].a[12]-'0');
    	}
    	sort(p,p+n,cmp1);
    	while(m--)
    	{
    		scanf("%d ",&x);		
    		if(x==1)
    		{
    			int f=0;
    			scanf("%c",&c);			
    			printf("Case %d: %d %c\n",o++,x,c);
    			for(i=0;i<n;i++)
    			{
    				if(p[i].type==c)
    				{
    					f=1;
    					cout<<p[i].a<<" "<<p[i].score<<endl;
    				}			
    			}
    			if(f==0)
    			printf("NA\n");
    		}
    		else if(x==2)
    		{
    			scanf("%d",&y);
    			printf("Case %d: %d %d\n",o++,x,y);
    			int s=0,sum=0;
    			for(i=0;i<n;i++)
    			{
    				if(p[i].site==y)
    				{
    					s++;
    					sum+=p[i].score;
    				}
    			}
    			if(s==0)
    			printf("NA\n");
    			else
    			printf("%d %d\n",s,sum);		
    		}
    		else if(x==3)
    		{
    			scanf("%d",&y);
    			int op[1000]={0},oo=0;
    			for(i=0;i<n;i++)
    			{
    				if(p[i].date==y)
    				{
    					op[p[i].site]++;
    				}
    			}
    			for(i=101;i<=999;i++)
    			{
    				if(op[i]!=0)
    				{
    					//cout<<i<<op[i]<<endl;
    					pp[oo].num=i;
    					pp[oo].sum=op[i];
    					oo++;
    				}
    			}
    			printf("Case %d: %d %06d\n",o++,x,y);
    			if(oo==0)
    			printf("NA\n");
    			else
    			{
    				sort(pp,pp+oo,cmp3);
    				for(i=0;i<oo;i++)
    				printf("%d %d\n",pp[i].num,pp[i].sum);
    			}
    			
    		}
    	}
    }