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类型1和2都好处理，就是3需要map，直接记录考场号和日期拼接字符串对应考生的个数，然后排序。代码：

#include <iostream>
#include <vector>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <map>
#define MAX 10001
using namespace std;
int n,m;
struct stu {
    char Tid[14];
    int score;
}s[MAX];
char *Substr(char *a,int l,int r) {
    char *t = (char *)malloc(sizeof(char) * (r - l + 1));
    for(int i = l;i < r;i ++) {
        t[i - l] = a[i];
    }
    t[r - l] = 0;
    return t;
}
int snum[1000],sscore[1000];///考场对应考生人数  和 总分的数组
int lnum[3];
char *str[MAX];///记录考场和日期总串
int c;
map<string,int> mp;
bool cmp(const stu &a,const stu &b) {
    if(a.Tid[0] == b.Tid[0]) {
        if(a.score == b.score) return strcmp(a.Tid,b.Tid) < 0;
        return a.score > b.score;
    }
    return a.Tid[0] < b.Tid[0];
}
bool cmp1(const char *a,const char *b) {
    if(mp[a] == mp[b]) return strcmp(a,b) < 0;
    return mp[a] > mp[b];
}
int main() {
    char *t;
    scanf("%d%d",&n,&m);
    for(int i = 0;i < n;i ++) {
        scanf("%s %d",s[i].Tid,&s[i].score);
        t = Substr(s[i].Tid,1,4);
        int d = atoi(t);
        snum[d] ++;
        sscore[d] += s[i].score;
        int dd = s[i].Tid[0] == ‘T‘ ? 2 : s[i].Tid[0] - ‘A‘;
        lnum[dd] ++;
        t = Substr(s[i].Tid,1,10);
        if(!mp[t]) {
            str[c] = (char *)malloc(sizeof(char) * 10);
            strcpy(str[c],t);
            c ++;
        }
        mp[t] ++;
    }
    sort(s,s + n,cmp);
    sort(str,str + c,cmp1);
    int choice,site;
    char level[2],date[7];
    for(int i = 1;i <= m;i ++) {
        scanf("%d",&choice);
        printf("Case %d: %d ",i,choice);
        if(choice == 1) {
            scanf("%s",level);
            printf("%s\n",level);
            int l,r;
            if(level[0] == ‘A‘) l = 0,r = lnum[0];
            else if(level[0] == ‘B‘) l = lnum[0],r = lnum[1];
            else l = lnum[0] + lnum[1],r = lnum[2];
            if(r == 0) {
                printf("NA\n");
                continue;
            }
            for(int j = 0;j < r;j ++) {
                printf("%s %d\n",s[l + j].Tid,s[l + j].score);
            }
        }
        else if(choice == 2) {
            scanf("%d",&site);
            printf("%d\n",site);
            if(!snum[site]) {
                printf("NA\n");
            }
            else {
                printf("%d %d\n",snum[site],sscore[site]);
            }
        }
        else {
            scanf("%s",&date);
            printf("%s\n",date);
            int d = 0;
            for(int j = 0;j < c;j ++) {
                if(strcmp(Substr(str[j],3,9),date) == 0) {
                    printf("%s %d\n",Substr(str[j],0,3),mp[str[j]]);
                    d ++;
                }
            }
            if(d == 0) printf("NA\n");
        }
    }
}

原文地址：https://www.cnblogs.com/8023spz/p/10211512.html