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 func…

1078 Hashing (25 分)   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 ( where TSize is the maximum size of the ha…

题目:https://pintia.cn/problem-sets/994805342720868352/problems/994805389634158592 题意: 给定哈希表的大小和n个数,使用平方探测法解决冲突,为每个数在哈希表中的位置. 如果给定的哈希表的大小不是质数,将其改为最小的比他大的质数. 思路: 比较基础的题目.有两个要注意的点! 1.初始化notprime数组时,需要注意1,也不是质数.特殊处理notprime[1] = true 2.注意考虑prob的边界,应该是哈希表的…

https://pintia.cn/problem-sets/994805342720868352/problems/994805389634158592 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 d…

题目 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…

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: 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(…

题意: 输入两个正整数M和N(M<=10000,N<=M)表示哈希表的最大长度和插入的元素个数.如果M不是一个素数,把它变成大于M的最小素数,接着输入N个元素,输出它们在哈希表中的位置(从0开始),如有冲突采取二次探测法处理冲突. trick: 测试点1包含M为1的数据,1不是素数... AAAAAccepted code: #include<bits/stdc++.h> using namespace std; ]; ]; ]; int main(){ ios::sync_wit…

1145 Hashing - Average Search Time (25 分)   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 ti…

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'-' #include <iostream> #include <cstdio> #include <algorithm> #include <string.h> #include <cmath> using namespace std; ; //之前设置…

二次探测法.表示第一次听说这东西... #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<stack> #include<queue> #include<string> #include<algorithm> using namespace std; ; bool fl…