题目 | Prefix Equality

AtCoder Beginner Contest 250

E - Prefix Equality

https://atcoder.jp/contests/abc250/tasks/abc250_e

Time Limit: 4 sec / Memory Limit: 1024 MB

Score : $500$ points

Problem Statement

You are given integer sequences $A=(a_1,\dots ,a_N)$ and $B=(b_1,\dots ,b_N)$, each of length NN.

For $i=1,...,Q$, answer the query in the following format.

• If the set of values contained in the first $x_i$ terms of $A$, $(a_1,\dots ,a_{x_i}),$ and the set of values contained in the first $y_i$ terms of $B$, $(b_1,\dots ,b_{y_i})$, are equal, then print Yes; otherwise, print No.

Constraints

• $1\le N,Q\le 2\times {10}^5$
• $1\le a_i,b_i\le {10}^9$
• $1\le x_i,y_i\le N$
• All values in input are integers.

---

Input

Input is given from Standard Input in the following format:

$N$
$a_1$ $\dots$ $a_N$
$b_1$ $\dots$ $b_N$
$Q$
$x_1$ $y_1$
$⋮$
$x_Q$ $y_Q$

Output

Print $Q$ lines. The $i$-th line should contain the response to the $i$-th query.

---

Sample Input 1

5
1 2 3 4 5
1 2 2 4 3
7
1 1
2 2
2 3
3 3
4 4
4 5
5 5

Sample Output 1

Yes
Yes
Yes
No
No
Yes
No

Note that sets are a concept where it matters only whether each value is contained or not.
For the $3$-rd query, the first $2$ terms of $A$ contain one $1$ and one $2$, while the first $3$ terms of $B$ contain one $1$ and two $2$'s. However, the sets of values contained in the segments are both $\{1,2\}$, which are equal.
Also, for the 66-th query, the values appear in different orders, but they are still equal as sets.

我的笔记

使用 $hash$ 计算出 $a_1\sim a_x$ 和 $b_1\sim b_x(1\le x\le N)$ 所含数字集合的 $hash$ 值，存到长为 $N$ 的数组 $hash\mathrm{\_}a$、$hash\mathrm{\_}b$ 内。

判断时，只需判断 $hash\mathrm{\_}a[x_i]$ 是否等于 $hash\mathrm{\_}b[y_i]$。若等于，则集合相等；不等，则集合不等。

$hash$ 函数可用 $f(x)=x\cdot (x+1326)\cdot (x+9185)$，使用自然溢出法。

源码

#include <bits/stdc++.h>

using namespace std;

const int MAXN = 2e5 + 20;
int N, Q;
unsigned int hash_a[MAXN], hash_b[MAXN];

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> N;
    set<int> st;
    for (int i = 0; i < N; i++)
    {
        int tmp;
        cin >> tmp;
        if (st.find(tmp) == st.end())
        {
            st.insert(tmp);
            hash_a[i + 1] = (hash_a[i] + tmp * (tmp + 1326) * (tmp + 9185));
        }
        else
        {
            hash_a[i + 1] = hash_a[i];
        }
    }
    st.clear();
    for (int i = 0; i < N; i++)
    {
        int tmp;
        cin >> tmp;
        if (st.find(tmp) == st.end())
        {
            st.insert(tmp);
            hash_b[i + 1] = (hash_b[i] + tmp * (tmp + 1326) * (tmp + 9185));
        }
        else
        {
            hash_b[i + 1] = hash_b[i];
        }
    }
    cin >> Q;
    while (Q--)
    {
        int a, b;
        cin >> a >> b;
        if (hash_a[a] == hash_b[b])
            cout << "Yes" << endl;
        else
            cout << "No" << endl;
    }
    return 0;
}