1153 Decode Registration Card of PAT (25 point(s))
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
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
本文參考了大佬:https://blog.csdn.net/Hickey_Chen/article/details/84933814
題目大意:N個學生考號,M個問題。題目讀懂了也不難,此處就不翻譯了,總結一下我遇到的問題吧:開始我把考號中的每一項都寫進結構體裡,以為處理起來方便點,超時了,之後放一起多了4分。之後便是最坑的cin,cout所需要的時間遠超過printf和scanf。改了後19。最後一點就是我將unordered_map定義在了全域性變數。。。。。寫在裡邊後滿分了。最後感謝大佬程式碼參考。程式碼如下:
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <unordered_map>
#include <algorithm>
using namespace std;
const int maxn=1e4+10;
struct card{
string id;
int score;
}St[maxn],co[maxn];
struct cr{
int cnt;
string tnum;
}Cr[maxn];
bool cmp(card a, card b)
{
return (a.score==b.score?a.id<b.id:a.score>b.score);
}
bool cmmp(cr a, cr b)
{
return (a.cnt==b.cnt?a.tnum<b.tnum:a.cnt>b.cnt);
}
int main()
{
// freopen("1153.txt","r",stdin);
int n,m,sc,type;
string str,s;
cin>>n>>m;
for(int i=0;i<n;i++){
cin>>str>>sc;
St[i].id=str;
St[i].score=sc;
}
for(int i=1;i<=m;i++){
cin>>type>>s;
int found=0;
switch(type){
case 1:
{
int k=0;
for(int j=0;j<n;j++){
char L=St[j].id[0];
if(L==s[0]){ //將符合條件的拷貝出去
co[k].id=St[j].id;
co[k].score=St[j].score;
k++;
found=1;
}
}
sort(co,co+k,cmp);
// cout<<"Case "<<i<<": "<<type<<" "<<s<<endl;
printf("Case %d: %d %s\n",i,type,s.c_str());
if(!found) printf("NA\n");//cout<<"NA\n";
else
for(int j=0;j<k;j++)
// cout<<co[j].id<<" "<<co[j].score<<endl;
printf("%s %d\n",co[j].id.c_str(),co[j].score);
break;
}
case 2:
{
int cnt=0,sum=0;
for(int j=0;j<n;j++){
// string s1=St[j].testnum;
string s1=St[j].id.substr(1,3);
if(s1==s){
cnt++;
sum+=St[j].score;
found=1;
}
}
// cout<<"Case "<<i<<": "<<type<<" "<<s<<endl;
printf("Case %d: %d %s\n",i,type,s.c_str());
if(!found) printf("NA\n");//cout<<"NA\n";
else
// cout<<cnt<<" "<<sum<<endl;
printf("%d %d\n",cnt,sum);
break;
}
case 3:
{
int z=0;
unordered_map<string,int> mp;
for(int j=0;j<n;j++){
string s1=St[j].id.substr(4,6);
string num1=St[j].id.substr(1,3);
if(s1==s){
mp[num1]++;
found=1;
}
}
// cout<<"Case "<<i<<": "<<type<<" "<<s<<endl;
printf("Case %d: %d %s\n",i,type,s.c_str());
if(!found) printf("NA\n");//cout<<"NA\n";
else{
for(auto it=mp.begin();it!=mp.end();it++){
string key=it->first;
int value=it->second;
Cr[z].tnum=key; //將結果存放到新的結構體中進行處理
Cr[z].cnt=value;
z++;
}
sort(Cr,Cr+z,cmmp);
for(int p=0;p<z;p++){
// cout<<Cr[p].tnum<<" "<<Cr[p].cnt<<endl;
printf("%s %d\n",Cr[p].tnum.c_str(),Cr[p].cnt);
}
}
break;
}
default:
// cout<<"Case "<<i<<": "<<type<<" "<<s<<endl<<"NA\n";
printf("Case %d: %d %s\nNA\n",i,type,s.c_str());
}
}
return 0;
}