题目描述

输入样例

5
I 1
I 2
I 3
Q 2
Q 5

输出样例

Yes
No

参考代码

拉链法

#include <iostream>
#include <cstring>
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;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 <iostream>
#include <cstring>
using namespace std;const int N = 200003, 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;scanf("%d", &n);memset(h, 0x3f, sizeof h);	// 每一个字节都是0x3f，int类型是4字节，所以是0x3f3f3f3fwhile (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;
}