# HDU 6038 Function （数学）

## Description

You are given a permutation a from 0 to n−1 and a permutation b from 0 to m−1.

Define that the domain of function f is the set of integers from 0 to n−1, and the range of it is the set of integers from 0 to m−1.

Please calculate the quantity of different functions f satisfying that f(i)=bf(ai) for each i from 0 to n−1.

Two functions are different if and only if there exists at least one integer from 0 to n−1 mapped into different integers in these two functions.

The answer may be too large, so please output it in modulo 10^9+7.

## Input

The input contains multiple test cases.

For each case:

The first line contains two numbers n, m. (1≤n≤100000,1≤m≤100000)

The second line contains n numbers, ranged from 0 to n−1, the i-th number of which represents ai−1.

The third line contains m numbers, ranged from 0 to m−1, the i-th number of which represents bi−1.

It is guaranteed that ∑n≤106, ∑m≤106.

## Output

For each test case, output “Case #x: y” in one line (without quotes), where x indicates the case number starting from 1 and y denotes the answer of corresponding case.

## Sample Input

3 2
1 0 2
0 1
3 4
2 0 1
0 2 3 1


## Sample Output

Case #1: 4
Case #2: 4


## 思路

$f(0)=b_{f(1)}$

$f(1)=b_{f(0)}$

$f(2)=b_{f(2)}$

## AC 代码

#include<iostream>
#include<vector>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdio>
using namespace std;

typedef __int64 LL;
const int maxn=110000;
const int mod = 1e9+7;

int a[maxn],b[maxn];
vector<int> na,nb;
bool vis[maxn];

void findnum(int *a,int n,vector<int> &res)     //寻找 res 中的循环节个数以及长度
{
memset(vis,false,sizeof(vis));
for(int i=0; i<n; i++)
{
if(!vis[i])
{
int now=a[i],len=0;
while(!vis[now])
{
++len;
vis[now]=true;
now=a[now];
}
res.push_back(len);
}
}
}

int main()
{
ios::sync_with_stdio(false);
int n,m,ti=0;
while(cin>>n>>m)
{
na.clear();
nb.clear();
for(int i=0; i<n; i++)
cin>>a[i];
for(int i=0; i<m; i++)
cin>>b[i];
findnum(a,n,na);
findnum(b,m,nb);
LL ans=1;
int lena=na.size();
int lenb=nb.size();
for(int i=0; i<lena; i++)
{
LL res=0;
for(int j=0; j<lenb; j++)
{
if(na[i]%nb[j]==0)
res=(res+nb[j])%mod;
}
ans=(ans*res)%mod;
}
cout<<"Case #"<<++ti<<": "<<ans<<endl;
}
return 0;
}