PAT 1078. Hashing
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
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 1078 Hashing (25 分)
1078 Hashing (25 分) The task of this problem is simple: insert a sequence of distinct positive integ ...
- 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'-' ...
- 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< ...
- 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
- 实现目标文件与源码分开的makefile测试实验
uboot提供了两种编译策略,即可以将生成的目标文件与源码混在一起,也可以将生成的目标文件与源码分开.通过对uboot Makefile的分析,笔者编写了一个简单的实现这种功能的Makfile. 顶层 ...
- delphi xe5 android sample
安装xe5以后demo存放的路径在 C:\users\Public\Documents\RAD Studio\12.0\Samples 另外易博龙在sourceforget上也有 svn地址为:sv ...
- C++的类和对象
#include <iostream> // 预处理命令 using namespace std; class Student{ // 声明一个类,类名为Student private : ...
- Discuz! 7.2 SQL注入exp
已经有人写出一些工具了,但是感觉不怎么好用,就自己写了个. 参数:1.可直接getshell2.爆管理账号密码3.爆表前缀如果表前缀不是默认的cdb_ 只需更改代码中的 $table即可,方便快捷. ...
- uva 12097 - Pie
简单题,二分就行: #include<cstdio> #include<cmath> #define pi acos(-1.0) #define eps 0.000001 #d ...
- [topcoder]CorrectMultiplicationTwo
http://community.topcoder.com/stat?c=problem_statement&pm=11609&rd=14547 http://apps.topcode ...
- win10全系列官方MSDN原版系统安装密钥、版本区别、镜像下载地址与激活教程
微软发布win10正式版已经过去几天了,相信很多同学都心情高涨的装上了期待已久的win10系统,但也有很多同学面对win10系统的众多版本感到一脸茫然,的确,微软在win10系统版本上的划分确实有点多 ...
- Hadoop开发环境搭建
hadoop是一个分布式系统基础架构,由Apache基金会所开发. 用户可以在不了解分布式底层细节的情况下,开发分布式程序.充分利用集群的威力高速运算和存储. Hadoop实现了一个分布式文件系统 ...
- Specify a culture in string conversion explicitly
Specify a culture in string conversion explicitly There are different methods of grouping symbols, l ...
- Why Creating a Meaningful Morning Routine Will Make You More Successful
https://medium.com/life-learning/how-creating-a-meaningful-morning-routine-will-make-you-more-succes ...