题目描述

思路讲解

代码展示

拉链法：

#include <cstring>
#include <iostream>using namespace std;const int N = 100003;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++;
}bool find(int x) {int k = (x % N + N) % N;  //让余数变成正数for (int i = h[k]; i != -1; i = ne[i])if (e[i] == x)return true;return false;
}int main() {int n;scanf("%d", &n);memset(h, -1, sizeof h);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 <cstring>
#include <iostream>using namespace std;const int N = 200003, null = 0x3f3f3f3f;int h[N];int find(int x) {int t = (x % N + N) % N;while (h[t] != null && h[t] != x) {t++;if (t == N) t = 0;}return t;
}int main() {memset(h, 0x3f, sizeof h);int n;scanf("%d", &n);while (n--) {char op[2];int x;scanf("%s%d", op, &x);if (*op == 'I') h[find(x)] = x;else {if (h[find(x)] == null) puts("No");else puts("Yes");}}return 0;
}