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
- 思路:有点墨迹,就是排序
对于 type 1:
先整体排序,按cmp_stu:分数 -> id
查询时,遍历所有学生,每遍历到相应查询等级就输出
对于 type 2:
输入时就用结构体数组sch[]
记录好每个考场的人数和总分数
对于 type 3:
有点麻烦,每次建一个Sch的数组,用来统计查询生日的学生 所在考场的人数,统计完排序输出
【注意】每次都要重置统计数组
-
坑点:
Wrong 1:样例1 3 WA ,4 TLE :动态查询必须每次更新 , 只能用scanf和printf输入输出 -
code:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 10010;
struct Stu{
string id;
int score;
}stu[maxn];
struct Sch{
int id, num, total_score;
Sch(){
num = total_score = 0;
}
}sch[maxn];
bool cmp_stu(Stu a, Stu b){
return a.score != b.score ? a.score > b.score : a.id < b.id;
}
bool cmp_sch(Sch a, Sch b){
return a.num != b.num ? a.num > b.num : a.id < b.id;
}
int main(){
int n, nq;
scanf("%d %d", &n, &nq);
for(int i = 0; i < n; ++i){
cin >> stu[i].id;
scanf("%d", &stu[i].score);
int t_site = stoi(stu[i].id.substr(1, 3));
sch[t_site].num++;
sch[t_site].total_score += stu[i].score;
}
sort(stu, stu + n, cmp_stu);
for(int i = 1; i <= nq; ++i){
int type;
char tmp[15];
scanf("%d %s", &type, tmp);
string term = tmp;
printf("Case %d: %d %s\n", i, type, tmp);
bool flg = true;
if(type == 1){
for(int i = 0; i < n; ++i){
if(stu[i].id[0] == term[0]){
printf("%s %d\n", stu[i].id.c_str(), stu[i].score);
flg = false;
}
}
}else if(type == 2){
int t_site = stoi(term);
if(sch[t_site].num != 0){
printf("%d %d\n", sch[t_site].num, sch[t_site].total_score);
flg = false;
}
}else{
vector<Sch> cnt(1000); //Wrong 1: 样例1 3 WA ,4 TLE 动态查询必须每次更新 , 只能用scanf和printf输入输出
for(int i = 0; i < n; ++i){
if(term == stu[i].id.substr(4, 6)){
int t_site = stoi(stu[i].id.substr(1, 3));
cnt[t_site].id = t_site;
cnt[t_site].num++;
}
}
sort(cnt.begin(), cnt.end(), cmp_sch);
for(int i = 0; cnt[i].num > 0 && i < cnt.size(); ++i){
printf("%d %d\n", cnt[i].id, cnt[i].num);
flg = false;
}
}
if(flg) printf("NA\n");
}
return 0;
}