The task of this problem is simple: insert a sequence of distinct positive integers into a hash table, and output the positions of the input numbers.  The hash function is defined to be "H(key) = key % TSize" where TSize is the maximum size of the hash table.  Quadratic probing (with positive increments only) is used to solve the collisions.

Note that the table size is better to be prime.  If the maximum size given by the user is not prime, you must re-define the table size to be the smallest prime number which is larger than the size given by the user.

Input Specification:

Each input file contains one test case.  For each case, the first line contains two positive numbers: MSize (<=104) and N (<=MSize) which are the user-defined table size and the number of input numbers, respectively.  Then N distinct positive integers are given in the next line.  All the numbers in a line are separated by a space.

Output Specification:

For each test case, print the corresponding positions (index starts from 0) of the input numbers in one line.  All the numbers in a line are separated by a space, and there must be no extra space at the end of the line.  In case it is impossible to insert the number, print "-" instead.

Sample Input:

4 4
10 6 4 15

Sample Output:

0 1 4 -
// 1078pat.cpp : 定义控制台应用程序的入口点。
// #include <iostream>
#include <string.h>
#include <vector>
using namespace std; const int N=;
int primes[N]; //找出所有的质数
bool mark[N]; //标记所有的质数
int res[N];
bool hashs[N];
vector<int> input;
int primeSize; void init()
{
memset(mark,false,N*sizeof(bool));
for(int i=;i<N;++i)
res[i]=-;
primeSize=;
for(int i=;i<N;++i)
{
if(mark[i]) continue;
primes[primeSize++]=i;
if(i>=) continue;
for(int j=i*i;j<N;j+=i)
{
mark[j]=true;
}
}
}
int main()
{
init();
int Msize,n;
cin>>Msize>>n;
if(Msize<n) //保证Msize<n;
Msize=n;
if(Msize<) //保证Msize==1时,令Msize=2;
Msize=;
if(mark[Msize])
{
for(int i=;i<N;++i)
{
if(primes[i]>Msize)
{
Msize=primes[i];
break;
}
}
}
int buf;
int position;
for(int i=;i<n;++i)
{
cin>>buf;
input.push_back(buf);
position=buf%Msize;
for(int j=position,k=;k<Msize;++k,j+=k*k)
//quadratic probing 0<=k<Msize
{
int newp=j%Msize;
if(!hashs[newp])
{
hashs[newp]=true;
res[buf]=newp;
break;
}
j=position;
}
}
for(int i=;i<n;++i)
{
if(i!=)
cout<<" ";
if(-==res[input[i]])
{
cout<<"-";
}
else
cout<<res[input[i]];
}
return ;
}

PAT 1078. Hashing的更多相关文章

  1. PAT 1078 Hashing[一般][二次探查法]

    1078 Hashing (25 分) The task of this problem is simple: insert a sequence of distinct positive integ ...

  2. 1078. Hashing (25)【Hash + 探測】——PAT (Advanced Level) Practise

    题目信息 1078. Hashing (25) 时间限制100 ms 内存限制65536 kB 代码长度限制16000 B The task of this problem is simple: in ...

  3. pat 甲级 1078. Hashing (25)

    1078. Hashing (25) 时间限制 100 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 task of t ...

  4. PAT 1145 1078| hashing哈希表 平方探测法

    pat 1145: 参考链接 Quadratic probing (with positive increments only) is used to solve the collisions.:平方 ...

  5. PAT 甲级 1078 Hashing (25 分)(简单,平方二次探测)

    1078 Hashing (25 分)   The task of this problem is simple: insert a sequence of distinct positive int ...

  6. 1078 Hashing (25 分)

    1078 Hashing (25 分) The task of this problem is simple: insert a sequence of distinct positive integ ...

  7. PAT甲题题解-1078. Hashing (25)-hash散列

    二次方探测解决冲突一开始理解错了,难怪一直WA.先寻找key%TSize的index处,如果冲突,那么依此寻找(key+j*j)%TSize的位置,j=1~TSize-1如果都没有空位,则输出'-' ...

  8. PAT (Advanced Level) 1078. Hashing (25)

    二次探测法.表示第一次听说这东西... #include<cstdio> #include<cstring> #include<cmath> #include< ...

  9. PAT甲级1078 Hashing【hash】

    题目:https://pintia.cn/problem-sets/994805342720868352/problems/994805389634158592 题意: 给定哈希表的大小和n个数,使用 ...

随机推荐

  1. pdo如何防止 sql注入

    我们使用传统的 mysql_connect .mysql_query方法来连接查询数据库时,如果过滤不严,就有SQL注入风险,导致网站被攻击,失去控制.虽然可以用 mysql_real_escape_ ...

  2. Leaflet学习笔记-Leaflet.awesome-markers

    基础篇传送门 http://www.cnblogs.com/CoffeeEddy/p/4919987.html 效果图 是不是感觉很美观啊 为什么选择awesome 地图上面需要各种各样的Mark,难 ...

  3. ListToDataTable

    public static DataTable ToDataTable<T>(IEnumerable<T> collection)       {           var ...

  4. HTML -- 元素和属性

    HTML -- 元素 HTML元素是从开始标签到结束标签之间的代码,如: <!-- 加粗标签 --> <b>一些元素</b> <!-- 换行 --> & ...

  5. Array 原型扩展(快速排序,搅乱顺序)

    /// 快速快速排序算法Array.prototype.quickSort = function (left, right) { // left = left || 0; // right = rig ...

  6. cmd下运行java文件时,找不到或无法加载主类的解决方法

    最近再看java,却被一个看似很基础的hellorworld头疼了十几分钟,百度了一下,若你在cmd下编辑及运行.java文件时报错,解决方案如下,如图所示:

  7. ORACLE 常用系统函数

    1.  字符类 1.1  ASCII(c ) 函数  和CHR( i )      ASCII 返回一个字符的ASCii码,其中c表示一个字符;CHR 返回ascii码值i 所对应的字符 . 如: S ...

  8. 在windows下创建.gitignore文件

    1.使用另存为的方式   2.在win7下,文件名输入 ”.gitignore.“ http://hbiao68.iteye.com/blog/2055496 http://lyhopq.github ...

  9. Introduction to Gaussian Processes

    Introduction to Gaussian Processes Gaussian processes (GP) are a cornerstone of modern machine learn ...

  10. XSS解决方案系列之四:关于编码

    本文准备说明以下几个问题: 1. 关于重复编码的问题 2. 关于编码的多种形式的问题 3. 关于编码的几个常见问题 [说明] 本文所述编码是指encode,可以理解为转义,而不是编程序写代码. 编码或 ...