題目
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
雞賊之處
1.准考證帶字母
只能用string存准考證,所以還得弄個結構體,存准考證和分數
2.要求排序
就因爲這個排序,一定要用結構體和vector,在自定義一個排序函數
3.收集、計算
如果是收集(查詢1),那隻用結構體和vector沒有問題。
如果是計算(查詢3),因爲結構體和vector的成員值是隻讀的(我試過了,成員值只能局部改,出了框改不了),只能用map。而map會超時(很慘了),所以只能用unodered_map了(頭文件同名,別忘了加)。
代碼註釋
1.substr
字符串截斷,PAT常見操作
2.for(auto i:v)
如果用不到STL容器的編號就可以用這個c++14的枚舉特性
滿分代碼
#include<iostream>
#include<vector>
#include<unordered_map>
using namespace std;
int n,m,k,num=0,total=0;
string s;
struct node{
string id;
int value;
};
bool cmp(node &a,node &b){
return a.value==b.value?a.id<b.id:a.value>b.value;
}
int main(){
scanf("%d %d",&n,&m);
vector<node> v(n);
for(int i=0;i<n;i++){
cin>>v[i].id>>v[i].value;
}
for(int i=1;i<=m;i++){
cin>>k>>s;
printf("Case %d: %d %s\n",i,k,s.c_str());
vector<node> ans;
if(k==1){
for(auto it:v){
if(it.id.substr(0,1)==s){
ans.push_back(it);
}
}
if(ans.size()==0){
printf("NA\n");
continue;
}
sort(ans.begin(),ans.end(),cmp);
for(auto j:ans){
printf("%s %d\n",j.id.c_str(),j.value);
}
}else if(k==2){
num=0;
total=0;
for(auto it:v){
if(it.id.substr(1,3)==s){
num++;
total+=it.value;
}
}
if(num==0){
printf("NA\n");
continue;
}
printf("%d %d\n",num,total);
}else if(k==3){
unordered_map<string, int> ans3;
for(auto it:v){
if(it.id.substr(4,6)==s){
ans3[it.id.substr(1,3)]++;
}
}
if(ans3.size()==0){
printf("NA\n");
continue;
}
for(auto it:ans3){
ans.push_back({it.first,it.second});
}
sort(ans.begin(),ans.end(),cmp);
for(auto j:ans){
printf("%s %d\n",j.id.c_str(),j.value);
}
}
}
return 0;
}
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