2017 Multi-University Training Contest

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Function

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1289    Accepted Submission(s): 591

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

a[]的排列确定出f[i]的递推关系，构成几个递推环，b[]的排列构成答案环，如果b[]中的环长度能作为a[]环的因子，那么b[]环作为结果代入，就没有矛盾

#include<iostream>#include<cstdio>#include<cstring>#include<vector>#include<algorithm>using namespace std;const long long MOD= 1e9+7;int n,m;int a[111111],b[111111];int f[111111];vector<int>Va,Vb;int ncase=0;int main(){    while(scanf("%d%d",&n,&m)==2)    {        Va.clear();Vb.clear();        for(int i=0;i<n;i++)scanf("%d",&a[i]);        for(int i=0;i<m;i++)scanf("%d",&b[i]);        memset(f,-1,sizeof(f));        for(int i=0;i<n;i++)        {            if(f[a[i]]==-1)            {                //f[a[i]]=0;                int cnt=0;                int k=a[i];                while(f[a[k]]==-1)                {                    f[a[k]]=0;                    k=a[k];                    cnt++;                }                if(cnt){Va.push_back(cnt);}            }        }        //for(int i=0;i<Va.size();i++)cout<<Va[i]<<" ";        memset(f,-1,sizeof(f));        for(int i=0;i<m;i++)        {            if(f[b[i]]==-1)            {                //f[a[i]]=0;                int cnt=0;                int k=b[i];                while(f[b[k]]==-1)                {                    f[b[k]]=0;                    k=b[k];                    cnt++;                }                if(cnt){Vb.push_back(cnt);}            }        }        sort(Va.begin(),Va.end());        sort(Vb.begin(),Vb.end());        int lena=Va.size();        int lenb=Vb.size();        long long ans=1;        int  flag=0;        for(int i=0;i<lena;i++)        {            long long mul=0;            for(int j=0;j<lenb;j++)            {                if(Va[i]<Vb[j])break;                if(Va[i]%Vb[j]==0){mul+=Vb[j];}            }            if(mul){flag++;ans=ans*mul%MOD;}            else {ans=0;break;}        }        //if(flag<lena)ans=0;        printf("Case #%d: %lld\n",++ncase,ans);    }    return 0;}