310651: CF1866B. Battling with Numbers

Memory Limit:256 MB Time Limit:1 S
Judge Style:Text Compare Creator:
Submit:0 Solved:0

Description

B. Battling with Numberstime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output

On the trip to campus during the mid semester exam period, Chaneka thinks of two positive integers $X$ and $Y$. Since the two integers can be very big, both are represented using their prime factorisations, such that:

  • $X=A_1^{B_1}\times A_2^{B_2}\times\ldots\times A_N^{B_N}$ (each $A_i$ is prime, each $B_i$ is positive, and $A_1<A_2<\ldots<A_N$)
  • $Y=C_1^{D_1}\times C_2^{D_2}\times\ldots\times C_M^{D_M}$ (each $C_j$ is prime, each $D_j$ is positive, and $C_1<C_2<\ldots<C_M$)

Chaneka ponders about these two integers for too long throughout the trip, so Chaneka's friend commands her "Gece, deh!" (move fast) in order to not be late for the exam.

Because of that command, Chaneka comes up with a problem, how many pairs of positive integers $p$ and $q$ such that $\text{LCM}(p, q) = X$ and $\text{GCD}(p, q) = Y$. Since the answer can be very big, output the answer modulo $998\,244\,353$.

Notes:

  • $\text{LCM}(p, q)$ is the smallest positive integer that is simultaneously divisible by $p$ and $q$.
  • $\text{GCD}(p, q)$ is the biggest positive integer that simultaneously divides $p$ and $q$.
Input

The first line contains a single integer $N$ ($1 \leq N \leq 10^5$) — the number of distinct primes in the prime factorisation of $X$.

The second line contains $N$ integers $A_1, A_2, A_3, \ldots, A_N$ ($2 \leq A_1 < A_2 < \ldots < A_N \leq 2 \cdot 10^6$; each $A_i$ is prime) — the primes in the prime factorisation of $X$.

The third line contains $N$ integers $B_1, B_2, B_3, \ldots, B_N$ ($1 \leq B_i \leq 10^5$) — the exponents in the prime factorisation of $X$.

The fourth line contains a single integer $M$ ($1 \leq M \leq 10^5$) — the number of distinct primes in the prime factorisation of $Y$.

The fifth line contains $M$ integers $C_1, C_2, C_3, \ldots, C_M$ ($2 \leq C_1 < C_2 < \ldots < C_M \leq 2 \cdot 10^6$; each $C_j$ is prime) — the primes in the prime factorisation of $Y$.

The sixth line contains $M$ integers $D_1, D_2, D_3, \ldots, D_M$ ($1 \leq D_j \leq 10^5$) — the exponents in the prime factorisation of $Y$.

Output

An integer representing the number of pairs of positive integers $p$ and $q$ such that $\text{LCM}(p, q) = X$ and $\text{GCD}(p, q) = Y$, modulo $998\,244\,353$.

ExamplesInput
4
2 3 5 7
2 1 1 2
2
3 7
1 1
Output
8
Input
2
1299721 1999993
100000 265
2
1299721 1999993
100000 265
Output
1
Input
2
2 5
1 1
2
2 3
1 1
Output
0
Note

In the first example, the integers are as follows:

  • $X=2^2\times3^1\times5^1\times7^2=2940$
  • $Y=3^1\times7^1=21$

The following are all possible pairs of $p$ and $q$:

  • $p=21$, $q=2940$
  • $p=84$, $q=735$
  • $p=105$, $q=588$
  • $p=147$, $q=420$
  • $p=420$, $q=147$
  • $p=588$, $q=105$
  • $p=735$, $q=84$
  • $p=2940$, $q=21$

In the third example, the integers are as follows:

  • $X=2^1\times5^1=10$
  • $Y=2^1\times3^1=6$

There is no pair $p$ and $q$ that simultaneously satisfies $\text{LCM}(p,q)=10$ and $\text{GCD}(p,q)=6$.

Output

题目大意:
在学期中考试去校园的路上,Chaneka想到了两个正整数X和Y。由于这两个整数可能非常大,因此它们都使用质因数分解来表示,使得:

X=A_1^{B_1}×A_2^{B_2}×…×A_N^{B_N}(每个A_i都是质数,每个B_i都是正数,并且A_1
Y=C_1^{D_1}×C_2^{D_2}×…×C_M^{D_M}(每个C_j都是质数,每个D_j都是正数,并且C_1
Chaneka在旅途中一直在思考这两个整数,所以Chaneka的朋友命令她“快点!”以免考试迟到。

因此,Chaneka提出了一个问题,有多少对正整数p和q,使得LCM(p,q)=X和GCD(p,q)=Y。由于答案可能非常大,因此输出答案mod 998,244,353。

注意:

LCM(p,q)是同时被p和q整除的最小正整数。

GCD(p,q)是同时整除p和q的最大正整数。

输入数据格式:
第一行包含一个整数N(1≤N≤10^5)——X的质因数分解中的不同质数的数量。

第二行包含N个整数A_1,A_2,A_3,…,A_N(2≤A_1
第三行包含N个整数B_1,B_2,B_3,…,B_N(1≤B_i≤10^5)——X的质因数分解中的指数。

第四行包含一个整数M(1≤M≤10^5)——Y的质因数分解中的不同质数的数量。

第五行包含M个整数C_1,C_2,C_3,…,C_M(2≤C_1
第六行包含M个整数D_1,D_2,D_3,…,D_M(1≤D_j≤10^5)——Y的质因数分解中的指数。

输出数据格式:
一个整数,表示满足LCM(p,q)=X和GCD(p,q)=Y的正整数对p和q的数量,模数998,244,353。题目大意: 在学期中考试去校园的路上,Chaneka想到了两个正整数X和Y。由于这两个整数可能非常大,因此它们都使用质因数分解来表示,使得: X=A_1^{B_1}×A_2^{B_2}×…×A_N^{B_N}(每个A_i都是质数,每个B_i都是正数,并且A_1

加入题单

上一题 下一题 算法标签: