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 21 0 20 13 42 0 10 2 3 1

 

Sample Output

Case #1: 4Case #2: 4

 

题意

给出两个序列 a 和 b ，求满足 f(i)=bf(ai) 的函数个数。

 

思路

根据样例 1 我们可以得到：

f(0)=bf(1)

f(1)=bf(0)

f(2)=bf(2)

代换以后有 f(0)=bf(1)=bbf(0) ， f(1)=bf(0)=bbf(1) ，也就是说，在这一个循环节中，我们只要确定其中一个数，其余的数都可以通过它来表示出来。

考虑置换 a 的一个循环节，长度为 l ，那么有 f(i)=bf(ai)=bbf(aai)=b⋯bf(i)l times b 。

于是 f(i) 的值在置换 b 中所在的循环节的长度必须为 l 的因数。

而如果 f(i) 的值确定下来了，这个循环节的另外 l−1 个数的函数值也都确定下来了（刚刚所讨论的问题）。

因此满足条件的函数个数就是 ∏ki=1∑j|lij⋅calj ，其中 k 是置换 a 中循环节的个数， li 表示置换 a 中第 i 个循环节的长度， calj 表示置换 b 中长度为 j 的循环节的个数。

 

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;}