# 【题目】Prefix Equality

AtCoder Beginner Contest 250

E - Prefix Equality

Time Limit: 4 sec / Memory Limit: 1024 MB

Score : $500$ points

### Problem Statement

You are given integer sequences $A = (a_1,\ldots,a_N)$ and $B = (b_1,\ldots,b_N)$, 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 $A$, $(a_1,\ldots,a_{x_i}),$ and the set of values contained in the first $y_i$ terms of $B$, $(b_1,\ldots,b_{y_i})$, are equal, then print Yes; otherwise, print No.

### Constraints

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

### Input

Input is given from Standard Input in the following format:

$N$
$a_1$ $\ldots$ $a_N$
$b_1$ $\ldots$ $b_N$
$Q$
$x_1$ $y_1$
$\vdots$
$x_Q$ $y_Q$

### Output

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

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

$hash$ 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;
}