PAT 1145 Hashing - Average Search Time

原题连接：1145 Hashing - Average Search Time (25分)
参考的柳神博客：1145. Hashing – Average Search Time (25) – 甲级

题目：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（平方探测法） (with positive increments only) is used to solve the collisions（冲突）.

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 10​4​​. 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 10​5​​.

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

题目大意： 给定一个序列，用平方探测法解决哈希冲突，然后给出m个数字，如果这个数字不能够被插入就输出”X cannot be inserted.”，然后输出这m个数字的平均查找时间。

思路： 首先要找到比用户给的数大的最小素数作为真正的tsize。然后实现用平方探测法插入一个序列，再实现查找的功能。注意：按照平方探测法查找，如果哈希表上查到的位置为0（没有数），说明这个数已经不可能找到了，也要退出循环，结束查找。

代码：

#include 
#include 
#include 
using namespace std;

bool isprime(int n){
    if(n <= 1) return false;
    for(int i = 2; i * i <= n; i ++ )   //注意小于等于
        if(n % i == 0) return false;
    return true;
}

int main(){
    int tsize, n, m, a; //tsize用户定义的表的大小，n插入整数的数量，m查找键值的数量。
    scanf("%d%d%d", &tsize, &n, &m);
    
    while(!isprime(tsize)) tsize ++ ;
    //printf("%d\n", tsize);
    
    vector<int> hash(tsize);

    for(int i = 0; i < n; i ++ ){   //插入
        int flag = 0;
        scanf("%d", &a);
        for(int j = 0; j < tsize; j ++ ){
            int pos = (a + j * j) % tsize;
            if(hash[pos] == 0){
                hash[pos] = a;
                flag = 1;
                break;
            }
        }
        if(!flag) printf("%d cannot be inserted.\n", a);
    }
    
    int ans = 0;  
    for(int i = 0; i < m; i ++ ){   //查找
        scanf("%d", &a);
        for(int j = 0; j <= tsize; j ++ ){  //当如果查找了TSize次，都有数是由ans = TSize+1
            ans ++ ;
            int pos = (a + j * j) % tsize;
            if(hash[pos] == a || hash[pos] == 0){
                break;
            }
            //else ans = tsize + 1;
        }   
    }
    printf("%.1f", ans * 1.0 / m);
    
    return 0;
}

遇到的问题： 在过程中遇到了Segmentation Fault错误，这个错误是访问到了系统给这个程序内存之外的内存，错误原因是在scanf的时候忘了加&。