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 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 -

提交代码

 #include<cstdio>
#include<stack>
#include<algorithm>
#include<iostream>
#include<stack>
#include<set>
#include<map>
#include<cmath>
using namespace std;
void getprime(int &msize){
int i=msize,j;
if(msize==||msize==||msize==){
msize=;
return;
}
for(;;i++){
if(i%==){
continue;
}
int maxl=sqrt(msize*1.0);
for(j=;j<=maxl;j+=){
if(i%j==){
break;
}
}
if(j>maxl){
msize=i;
return;
}
}
}
map<int,int> ha;
bool g[];
int main(){
//freopen("D:\\INPUT.txt","r",stdin);
int i,msize,n,num;
scanf("%d %d",&msize,&n);
getprime(msize); //cout<<msize<<endl; for(i=;i<n;i++){
scanf("%d",&num);
ha[i]=num%msize;
}
int half=msize/;
map<int,int>::iterator it=ha.begin();
g[it->second]=true;
printf("%d",it->second);
it++;
for(;it!=ha.end();it++){
if(!g[it->second]){
g[it->second]=true;
printf(" %d",it->second);
}
else{
for(i=;i<=half;i++){
if(!g[(it->second+i*i)%msize]){
g[(it->second+i*i)%msize]=true;
printf(" %d",(it->second+i*i)%msize);
break;
}
}
if(i>half){
printf(" -");
}
}
}
printf("\n");
return ;
}

pat1078. Hashing (25)的更多相关文章

  1. PAT1078 Hashing

    11-散列2 Hashing   (25分) The task of this problem is simple: insert a sequence of distinct positive in ...

  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. pat09-散列1. Hashing (25)

    09-散列1. Hashing (25) 时间限制 100 ms 内存限制 65536 kB 代码长度限制 8000 B 判题程序 Standard 作者 CHEN, Yue The task of ...

  4. 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 ...

  5. PTA 11-散列2 Hashing (25分)

    题目地址 https://pta.patest.cn/pta/test/16/exam/4/question/679 5-17 Hashing   (25分) The task of this pro ...

  6. 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 ...

  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-1078 Hashing (散列表 二次探测法)

    1078. Hashing The task of this problem is simple: insert a sequence of distinct positive integers in ...

  9. 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 this problem is simp ...

随机推荐

  1. Java中使用同步关键字synchronized需要注意的问题

    在Java中,synchronized关键字是用来控制线程同步的,就是在多线程的环境下,控制synchronized代码段不被多个线程同时执行.synchronized既可以加在一段代码上,也可以加在 ...

  2. 滑动swipe的妙用

    转自:http://www.cnblogs.com/NEOCSL/archive/2013/03/04/2942861.html iterface ITouchable; function OnPic ...

  3. TS学习之基础类型

    1.布尔值 let isDone:boolean = false 2.数字(支持二,八,十,十六进制) let width:number = 20 3.字符串 let name:string = &q ...

  4. javaScript之事件处理程序

    事件就是用户或浏览器自身执行的某个动作,JavaScript与HTML的交互也是通过事件实现的.而相应某个事件的函数就叫做事件处理函数.包括以下几种: 1.HTML事件处理程序    某个元素支持的每 ...

  5. Dexdump 无法正常反编译问题

    WIN环境下无法正常运行,提示Unable open XXX as zip 解决方案:使用APKTOOL + JD-GUI进行替代反编译

  6. 【总结整理】关于Json的解析,校验和验证

    var jasondata='{"staff": [{"name":"红旗","age":90}, {"nam ...

  7. 错误:Tomcat version 7.0 only supports J2EE 1.2, 1.3, 1.4, and Java EE 5 and 6 Web

    在eclipse的workspace里面找到该项目. 依次进入:.settings->org.eclipse.wst.common.project.facet.core.xml. 打开文件后,将 ...

  8. hbase-0.98.1-cdh5.1.0 完全分布式搭建

    cdh版与0.98版的配置一样 1.环境 master:c1 slave:c2,c3 CentOS 6.5 x64 ,hadoop-2.3.0-cdh5.1.0,zookeeper-3.4.5-cdh ...

  9. Struts2学习第三课 Struts2详解

    接着上次的课程 这次我们看struts.xml 修改如下:这里是加上命名空间,默认的是不加,我们手动加上时就要在访问时加上命名空间. <?xml version="1.0" ...

  10. CSS, Sass, SCSS 关系

    Sass(Syntactically Awesome Style Sheets) ,是一种css预处理器和一种语言, 它可以用来定义一套新的语法规则和函数,以加强和提升CSS. 它有很多很好的特性,但 ...