PAT甲级——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 andB
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 correspondingTerm
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 correspondingTerm
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 correspondingTerm
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
whereNt
is the total number of testees andNs
is their total score; - for a type 3 query, output in the format
Site Nt
whereSite
is the site number andNt
is the total number of testees atSite
. The output must be in non-increasing order ofNt
's, or in increasing order of site numbers if there is a tie ofNt
.
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
这题....emmm, 用的柳婼的题解,一开始不太习惯这种题目。
//1 -->T,A,B
//2-4 -->考点 101-999
//5-10 -->data yymmdd
//11-13 -->考试数字 000-999
#include <iostream>
#include <vector>
#include <unordered_map>
#include <algorithm>
using namespace std;
struct node {
string t;
int value;
};
bool cmp(const node &a, const node &b) {
return a.value != b.value ? a.value > b.value : a.t < b.t; //三目运算符
}
int main() {
int n, k, num;
string s;
cin >> n >> k;
vector<node> v(n);
for (int i = 0; i < n; i++)
cin >> v[i].t >> v[i].value;
for (int i = 1; i <= k; i++) {
cin >> num >> s;
printf("Case %d: %d %s\n", i, num, s.c_str()); //c_str()将string转化为c语言的数组
vector<node> ans;
int cnt = 0, sum = 0;
if (num == 1) {
for (int j = 0; j < n; j++)
if (v[j].t[0] == s[0]) ans.push_back(v[j]);
}
else if (num == 2) {
for (int j = 0; j < n; j++) {
if (v[j].t.substr(1, 3) == s) {
cnt++;
sum += v[j].value;
}
}
if (cnt != 0) printf("%d %d\n", cnt, sum);
}
else if (num == 3) {
unordered_map<string, int> m;
for (int j = 0; j < n; j++)
if (v[j].t.substr(4, 6) == s) m[v[j].t.substr(1, 3)]++;
for (auto it : m) ans.push_back({it.first, it.second});
}
sort(ans.begin(), ans.end(),cmp);
for (int j = 0; j < ans.size(); j++)
printf("%s %d\n", ans[j].t.c_str(), ans[j].value);
if (((num == 1 || num == 3) && ans.size() == 0) || (num == 2 && cnt ==
0)) printf("NA\n");
}
return 0;
}
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