1145 Hashing - Average Search Time （25 point(s)）

The task of this problem is simple: insert a sequence of distinct positive integers into a hash table first. Then try to find another sequence of integer keys from the table and output the average search time (the number of comparisons made to find whether or not the key is in the table). 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 3 positive numbers: MSize, N, and M, which are the user-defined table size, the number of input numbers, and the number of keys to be found, respectively. All the three numbers are no more than 104. Then N distinct positive integers are given in the next line, followed by M positive integer keys in the next line. All the numbers in a line are separated by a space and are no more than 105.

Output Specification:

For each test case, in case it is impossible to insert some number, print in a line ​​X cannot be inserted.​​​ where ​​X​​ is the input number. Finally print in a line the average search time for all the M keys, accurate up to 1 decimal place.

Sample Input:

4 5 4
10 6 4 15 11
11 4 15 2

Sample Output:

15 cannot be inserted.
2.8

经验总结：

做过​​PAT 甲级 1078 Hashing​​​ 这一题的童鞋，遇到此题就比较轻松了。
平方探测的方法了，公式是newaddress=(address+step*step)%TableSize，这里传统观念可能会忘记对TableSize进行取余。
还有，step需要增长到TableSize才可以结束，这就是结束条件的控制，可以理论证明。
这里多了对元素进行查询的过程，实际上和插入元素十分类似（看代码就知道了），只不过停止条件是，当前访问的位置的元素为所查询的元素，或者为空时，跳出循环，不过还有一种情况是，step增长至TableSize仍然没有找到或者遇到空位置，此时查找次数最大，为TableSize+1。
输出时，精确到一位小数，就直接指定输出格式就可以了，无需四舍五入或者其他的操作。

AC代码

#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn=10010;
bool flag[maxn]={false};
int msize,n,m,prime[maxn],pnum=0,hash[maxn]={0},t;
void find_prime()
{
  flag[0]=flag[1]=true;
  for(int i=2;i<maxn;++i)
  {
    if(flag[i]==false)
    {
      prime[pnum++]=i;
      for(int j=i+i;j<maxn;j+=i)
        flag[j]=true;
    }
  }
}
int main()
{
  scanf("%d%d%d",&msize,&n,&m);
  find_prime();
  if(flag[msize]==true)
  {
    int *a=upper_bound(prime,prime+pnum,msize);
    msize=*a;
  }
  for(int i=0;i<n;++i)
  {
    bool f=false;
    scanf("%d",&t);
    int step=0;
    while(step<=msize)
    {
      int key=(t+step*step)%msize;
      if(hash[key]==0)
      {
        f=true;
        hash[key]=t;
        break;
      }
      ++step;
    }
    if(f==false)
      printf("%d cannot be inserted.\n",t);
  }
  int cnt=0;
  for(int i=0;i<m;++i)
  {
    bool f=false;
    scanf("%d",&t);
    int step=0;
    while(step<=msize)
    {
      int key=(t+step*step)%msize;
      if(hash[key]==0||hash[key]==t)
      {
        f=true;
        cnt+=step+1;
        break;
      }
      ++step;
    }
    if(f==false)
      cnt+=step;
  }
  printf("%.1f\n",cnt*1.0/m);
  return 0;
}