拉链法和开放寻址法对比 具体方法见 之前写的题解

拉链法：

#include <iostream>
#include <cstring>

using namespace std;

const int N = 100003; //!!!!
int e[N], h[N], ne[N], idx; //!!!!

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

bool find(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)); //!!!! 如果忘记初始化h[]会TLE！！！
    while (n --) {
        char op[2];
        int x;
        scanf("%s%d", op, &x);
        if (*op == 'I') insert(x);//!!!!
        else {
            if (find(x)) puts("Yes");
            else puts("No");
        }
    }
    return 0;
}

开放寻址法：

#include <iostream>
#include <cstring>

using namespace std;

const int N = 200003;//!!!!
const int 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;
        scanf("%s%d", op, &x);
        int k = find(x);
        if (*op == 'I') h[k] = x;//!!!!
        else {
            if (h[k] != null) puts("Yes");
            else puts("No");
        }
    }
    return 0;
}