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. rpc远程调用开发

    RPC即远程过程调用,适用于集群管理,集群节点就是RPCServer,而我们发起远程调用的web服务器就是RPCClient.所以是少数rpcClient(可能一个)对多个RPCServer(集群节点 ...

  2. javaScript & jquery完美判断图片是否加载完毕

    好久没写东西了,正好最近因为工作需要,写了一个瀑布流异步加载的程序. 今天就不谈瀑布流,来谈一下关于load的问题. ----------------------------------------- ...

  3. 翻译:ECMAScript 5.1简介

    简介 ECMAScript 5.1 (或仅 ES5) 是ECMAScript(基于JavaScript的规范)标准最新修正. 与HTML5规范进程本质类似,ES5通过对现有JavaScript方法添加 ...

  4. Python Tips and Traps(二)

    6.collections 模块还提供有OrderedDict,用于获取有序字典 import collections d = {'b':3, 'a':1,'x':4 ,'z':2} dd = col ...

  5. Python的数字类型及其技巧

    Python中的数字类型 int float fractions.Fraction decimal.Decimal 数字的舍与入 int(f):舍去小数部分,只保留整数部分,所以int(-3.8)的结 ...

  6. Hbase案例分析(一)

    Hbase应用场景: 1 随机读或者写 2 大数据上的高并发操作,比如每秒对PB级数据进行上千次操作.(查询,删除等操作) 3 读写均是非常简单的操作,比如没有join操作 Hbase Schema设 ...

  7. gdb调试高级用法

    Linux下进程崩溃时定位源代码位置 如何在调试内核时,同时可以调试应用程序的做法: (cskygdb) c Continuing. ^C Program received signal SIGINT ...

  8. Multi-Die系统介绍

    一个典型的存储系统一般是有几片NAND存储器组成的.一般会使用8-bit的总线,用来将不同的存储器与控制器进行连接,如图2.32所示.一个系统中多片NAND的存储系统可以提高存储容量,同时还可以提高读 ...

  9. Xamarin 开发常见问题

    原文:Xamarin 开发常见问题 Verify the project is selected to be deployed in the Solution Configuration Manage ...

  10. struts一点心得

    action中: 设置属性并增加get,set方法,给属性赋值后 (如: private String name; public String getName() { return name; } p ...