PAT甲级——A1129 Recommendation System【25】
Recommendation system predicts the preference that a user would give to an item. Now you are asked to program a very simple recommendation system that rates the user's preference by the number of times that an item has been accessed by this user.
Input Specification:
Each input file contains one test case. For each test case, the first line contains two positive integers: N (≤ 50,000), the total number of queries, and K (≤ 10), the maximum number of recommendations the system must show to the user. Then given in the second line are the indices of items that the user is accessing -- for the sake of simplicity, all the items are indexed from 1 to N. All the numbers in a line are separated by a space.
Output Specification:
For each case, process the queries one by one. Output the recommendations for each query in a line in the format:
query: rec[1] rec[2] ... rec[K]
where query is the item that the user is accessing, and rec[i] (i=1, ... K) is the i-th item that the system recommends to the user. The first K items that have been accessed most frequently are supposed to be recommended in non-increasing order of their frequencies. If there is a tie, the items will be ordered by their indices in increasing order.
Note: there is no output for the first item since it is impossible to give any recommendation at the time. It is guaranteed to have the output for at least one query.
Sample Input:
12 3
3 5 7 5 5 3 2 1 8 3 8 12
Sample Output:
5: 3
7: 3 5
5: 3 5 7
5: 5 3 7
3: 5 3 7
2: 5 3 7
1: 5 3 2
8: 5 3 1
3: 5 3 1
8: 3 5 1
12: 3 5 8
//题目大意:就是顾客买东西之前,根据顾客的购买历史进行商品推荐
//所以购买第一件商品3是没有推荐的,因为没有购买历史记录
//然后根据前面购买的历史次数,进行次数从多到少进行推荐,最多推荐三个,相同次数的推荐标号最小的的//
//使用set来得到按要求的排序规格,vector来辅助查询set中是否已经存在元素
#include <iostream>
#include <set>
#include <vector>
using namespace std;
int n, k;
struct Node
{
int value, cnt;
bool operator < (const Node& a)const//set的排序规则
{
return (cnt != a.cnt) ? cnt > a.cnt:value < a.value;
}
};
int main()
{
int n, k, num;
scanf("%d %d", &n, &k);
set<Node>s;
vector<int>v(n + , );
for (int i = ; i < n; ++i)
{
cin >> num;
if (i != )//第一件商品没有推荐
{
cout << num << ":";
int i = ;
for (auto ptr = s.begin(); ptr != s.end() && i <= k; ++ptr, i++)
cout << " " << ptr->value;
cout << endl;
}
auto ptr = s.find(Node{ num,v[num] });
if (ptr != s.end())//记住set不能修改内容,只能擦除后重新写入
s.erase(ptr);
v[num]++;
s.insert(Node{ num,v[num] });//使用了初始化列表原则
}
return ;
}
PAT甲级——A1129 Recommendation System【25】的更多相关文章
- PAT甲级 1129. Recommendation System (25)
1129. Recommendation System (25) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue ...
- PAT 甲级 1129 Recommendation System
https://pintia.cn/problem-sets/994805342720868352/problems/994805348471259136 Recommendation system ...
- PAT A1129 Recommendation System (25 分)——set,结构体重载小于号
Recommendation system predicts the preference that a user would give to an item. Now you are asked t ...
- A1129. Recommendation System
Recommendation system predicts the preference that a user would give to an item. Now you are asked t ...
- 1129. Recommendation System (25)
Recommendation system predicts the preference that a user would give to an item. Now you are asked t ...
- 【PAT甲级】1070 Mooncake (25 分)(贪心水中水)
题意: 输入两个正整数N和M(存疑M是否为整数,N<=1000,M<=500)表示月饼的种数和市场对于月饼的最大需求,接着输入N个正整数表示某种月饼的库存,再输入N个正数表示某种月饼库存全 ...
- PAT甲题题解-1129. Recommendation System (25)-排序
博主欢迎转载,但请给出本文链接,我尊重你,你尊重我,谢谢~http://www.cnblogs.com/chenxiwenruo/p/6789819.html特别不喜欢那些随便转载别人的原创文章又不给 ...
- PAT甲级 1122. Hamiltonian Cycle (25)
1122. Hamiltonian Cycle (25) 时间限制 300 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue The ...
- PAT甲级 1121. Damn Single (25)
1121. Damn Single (25) 时间限制 300 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue "Dam ...
随机推荐
- [转]WPF中的导航框架
有的时候,我们需要一个支持页面跳转的UI,例如文件浏览器,开始向导等.对于这样的界面,简单的可以使用ContentControl + ContentTemplateSelector的方式来实现,但是有 ...
- java常用类——比较器
Comparable和Comparator接口都是为了对类进行比较,众所周知,诸如Integer,double等基本数据类型,java可以对他们进行比较,而对于类的比较,需要人工定义比较用到的字段比较 ...
- css内容超出显示省略号
CSS实现单行.溢出显示省略号(…) 把要设置的显示省略号的标签,加上以下的属性 overflow: hidden; /*超出不显示*/ text-overflow: ellipsis;/* 超出内容 ...
- Linux命令(2):cat
cat命令是一个文本连接和查看的命令,用于文件的输出显示. 三大功能 一次显示整个文件. $ cat filename 从键盘创建一个文件.只能创建新文件,不能编辑已有文件. $ cat > f ...
- [转] boost undefined reference to 'pthread_create 问题
由于是Linux新手,所以现在才开始接触线程编程,照着GUN/Linux编程指南中的一个例子输入编译,结果出现如下错误:undefined reference to 'pthread_create'u ...
- Beaglebone Black的引脚分配
转自:http://blog.csdn.net/daxueba/article/details/50998000 Beaglebone Black的引脚分配 绝大多数的微型开发平台都提供了一些称为GP ...
- LOJ #113. 最大异或和 (线性基)
题目链接:#113. 最大异或和 题目描述 这是一道模板题. 给由 \(n\) 个数组成的一个可重集 \(S\),每次给定一个数 \(k\),求一个集合 \(T \subseteq S\),使得集合 ...
- 拾遗:Linux 用户及权限管理基础
Lacks of Knowledge 1: Linux has large amount of COMMANDS,but many of them have similar funtions,it's ...
- 剑指offer——28对称的二叉树
题目描述 请实现一个函数,用来判断一颗二叉树是不是对称的.注意,如果一个二叉树同此二叉树的镜像是同样的,定义其为对称的. 题解: 使用正常前序遍历与反向的前序遍历进行比较结果即可,注意,需将空 ...
- 创建一个学生表student,默认的表空间为users,字段自定,同时为表的各个字段分别添加合适的约束,然后测试约束的验证状态。
create table student(id number(4) constraint prim_key primary key,name varchar(8) not null,sex varch ...