题目链接：模拟散列表

拉链法

#include <iostream>
#include <cstring>

using namespace std;

const int N = 100010;

int h[N], e[N], ne[N], idx;


void insert(int x)
{
    int k = (x % N + N) % N;
    e[idx] = x;
    ne[idx] = h[k];
    h[k] = idx;
    idx ++;
}


bool query(int x)
{
    int k = (x % N + N) % N;
    for(int i = h[k]; i != -1; i = ne[i])
    {
        if(e[i] == x) return true;
    }
    return false;
    
}

int main()
{
    int n;
    cin >> n;
    memset(h, -1, sizeof h);
    while(n--)
    {
        char op[2];
        int x;
        cin >> op >> x;
        if(*op == 'I') insert(x);
        else
        {
            if(query(x)) cout << "Yes" << endl;
            else cout << "No" << endl;
        }
    }
    return 0;
}

开放寻址法

#include <iostream>
#include <cstring>

using namespace std;

const int N = 200010, null = 0x3f3f3f3f;
int h[N];

int find(int x)
{
    int k = (x % N + N) % N;
    while(h[k] != null && h[k] != x)
    {
        k ++;
        if(k == N) k = 0;
    }
    return k;
}

int main()
{
    int n;
    cin >> n;
    memset(h, 0x3f, sizeof h);
    
    while(n--)
    {
        char op[2];
        int x;
        cin >> op >> x;
        if(*op == 'I') h[find(x)] = x;
        else
        {
            if(h[find(x)] != null) cout << "Yes" << endl;
            else cout << "No" << endl;
        }
    }
    return 0;
}