2017多校1 1006Function

Function

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1530    Accepted Submission(s): 720

Problem 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 from0 to n−1 mapped into different integers in these two functions.

The answer may be too large, so please output it in modulo 109+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

 

给2组数求不同函数关系的种类，根据题目的意思，就是求2组数列中不同环的数量，因为根据多对一的关系，将第二个数组中将 是第一个数组中不用数量的环的因子数的环相加，再最终将结果相乘即可。可以用vector容器来存不同数量的环

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<cmath>#include<iomanip>#include<queue>#include<vector>using namespace std;const long long int MOD=7+1e9;vector<int> al,bl;int a[111111],b[111111];bool v[111111];int main(){    int n,m;    int cs=0;    ios::sync_with_stdio(false);    while(cin>>n>>m)    {        al.clear();        bl.clear();        int i;        for(i=0;i<n;i++)        cin>>a[i];        for(i=0;i<m;i++)        cin>>b[i];        memset(v,0,sizeof(v));        for(i=0;i<n;i++)        {            if(!v[i])            {                int now=a[i];                int len=0;                while(!v[now])                {                    v[now]=1;                    len++;                    now=a[now];                }                al.push_back(len);            }        }        memset(v,0,sizeof(v));        for(i=0;i<m;i++)        {            if(!v[i])            {                int now=b[i];                int len=0;                while(!v[now])                {                    v[now]=1;                    ++len;                    now=b[now];                }                bl.push_back(len);            }        }        int j;        long long int ans=1;        for(i=0;i<al.size();i++)        {            long long int ans1=0;            for(j=0;j<bl.size();j++)             if(al[i]%bl[j]==0)              ans1=(ans1+bl[j])%MOD;              ans=(ans*ans1)%MOD;        }        cout<<"Case #"<<++cs<<": ";        cout<<ans%MOD<<endl;    }    return 0;}