分解PAT字符串进行不同部分统计
使用cin,cout,string超时

1. 原文

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 (≤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 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

2. 解析

14位字符串分别提取不同部分进行统计

ABT级别 level
考场 site
年份 year

📝记录输入值 stu[i] –> node(char id[15],int grade) 数组

根据查询提取不同id部分

level: id[0]==level[0]

分数从大到小排，相同时按字母排 – sort
对应site的数量++,分数+=grade 存入map中
需要提交5-10位的year，2-4位的site –>map.vector

3. AC代码

#include<cstdio>
#include<map>
#include<vector>
#include<algorithm>
#include<cstring>
using namespace std;
char str[20]; int n;
map<int,int> cnt;
map<int,int> score;
struct node
{
	char id[20];
	int grade;
}stu[10010];
bool cmp(node a,node b){
	if(a.id[0]!=b.id[0]){
		return a.id[0]<b.id[0];
	}
	else if (a.grade!=b.grade)
	{
		return a.grade>b.grade;
	}else{
		return strcmp(a.id,b.id)<0;
	}
}
struct node2
{
	int site,cnt;
};
bool cmp2(node2 a,node2 b){
	if (a.cnt!=b.cnt)
	{
		return a.cnt>b.cnt;
	}else{
		return a.site<b.site;
	}
}
void dealevel(char c)
{
	int count=0;
	for (int i = 0; i < n; ++i)
	{
		if (stu[i].id[0]==c)
		{
			printf("%s %d\n",stu[i].id,stu[i].grade);
			count++;
		}
	}
	if (count==0)
	{
		printf("NA\n");
	}
}
void dealsite(char c[])
{
	int temp=0;
	for (int i = 0; i < 3; ++i)
	{
		temp=temp*10+c[i]-'0';
	}
	if (cnt[temp]!=0)
	{
		printf("%d %d\n",cnt[temp],score[temp]);
	}else{
		printf("NA\n");
	}
}
void dealyear(char c[])
{
	map<int,int> sites;
	vector<node2> vTemp;
	int year=0; int flag=0;
	for (int j = 0; j < 6; ++j)
	{
		year=year*10+c[j]-'0';
	}
	for (int i = 0; i < n; ++i)
	{
		int id=0; 
		for (int j = 4; j < 10; ++j)
		{
			id=id*10+stu[i].id[j]-'0';
		}
		//printf("%s %s\n",id.c_str(),year.c_str());
		if (id==year)
		{
			flag=1;
			int ans=0;
			for (int j = 1; j < 4; ++j)
			{
				ans=ans*10+stu[i].id[j]-'0';
			}
			sites[ans]++;
		}
	}
	if (flag==0)
	{
		printf("NA\n");
	}else{
		node2 temp;
		for (map<int,int>::iterator i = sites.begin(); i != sites.end(); ++i)
		{
			temp.site=i->first;
			temp.cnt=i->second;
			vTemp.push_back(temp);
		}
		sort(vTemp.begin(),vTemp.end(),cmp2);
		for (int i = 0; i < vTemp.size(); ++i)
		{
			printf("%d %d\n",vTemp[i].site,vTemp[i].cnt);
		}
	}
}
int main()
{
	int m;
	scanf("%d%d",&n,&m);
	node temp; 
	for (int i = 0; i < n; ++i)
	{
		scanf("%s%d",temp.id,&temp.grade);
		stu[i]=temp;
		int site=0;
		for (int j = 1; j < 4; ++j)
		{
			site=site*10+temp.id[j]-'0';
		}
		//printf("%s\n",site.c_str());
		cnt[site]++;
		score[site]+=temp.grade;
	}
	sort(stu,stu+n,cmp);
	int idx;
	for (int i = 1; i <= m; ++i)
	{
		scanf("%d%s",&idx,str);
		printf("Case %d: %d %s\n",i,idx,str);
		if (idx==1)
		{
			dealevel(str[0]);
		}else if (idx==2)
		{
			dealsite(str);
		}else if (idx==3)
		{
			dealyear(str);
		}else{
			printf("NA\n");
		}
	}
	return 0;
}

使用cin，cout，string 出现超时
使用char[] 数据的提取转为int存储

int a=0;
for (int j = 1; j < 4; ++j)
{
   a=a*10+id[j]-'0';
}

W.T.的博客

A1153 Decode Registration Card of PAT (25)

1. 原文

2. 解析

3. AC代码