Function

Function

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 1472    Accepted Submission(s): 691

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

  

题目大意：

       有a，b两个数组，它们之间满足f(i)=b(a(i))，求有多少种赋值方法。

思路：

       因为数组a可以与函数f的顺序构成关系，所以就把a固定，把b一一对应进去。这样就转化成一个循环问题，通过举例我们可以看出，我们需要先求出数组a与b的各个循环节的长度及个数，如果b的循环节长度不是a的循环节长度的因子，则不计入计算，我们只考虑b的循环节长度是a的循环节长度的因子的情况。最后结果等于所有为a循环节长度因子的b的循环节的长度乘以这个循环节的个数,再乘有多少这种循环节长度的循环。下面举例说明。

       7 6

       3 2 0 1 5 6 4

       1 0 3 4 2 5

       由此例可知，a的循环节有，(3 2 0 1)长度为4，个数为1；(5 6 4)长度为3，个数为1。b的循环节有，(1 0)长度为2，个数为1；(3 4 2)长度为3，个数为1；(5)长度为1，个数为1。其中1，2为4的因子；1，3为3的因子。则结果为(1*1+2*1)*(1*1+3*1)=3*4=12。

下面贴上代码：(具体实现过程及优化方法详解见代码注释)

#include<cstdio>#include<iostream>#include<cstring>#include<cmath>using namespace std;#define ll  long long#define Mod 1000000007const int maxn=1e5+5;ll circle[3][maxn]; //circle[a或者b][length]=个数int a[maxn],b[maxn];int main(){    int n,m;    int ca=1;    while(~scanf("%d %d",&n,&m)) //注意使用scanf,用cin会超时    {        memset(circle,0,sizeof circle);        memset(a,0,sizeof a);        memset(b,0,sizeof b);        for(int i=0; i<n; i++)            scanf("%d",&a[i]);        for(int i=0; i<n; i++)        {            ll k=i;            ll length=0;            while(a[k]!=-1) //若当前位置为-1,说明已经查找过,形成循环            {                length++; //length记录循环节的长度                int t=k; //t为当前查找的位置                k=a[k]; //根据a[位置]=下一个位置,查找下一个位置                a[t]=-1; //把查找到的位置标记为-1            }            if(length!=0)                circle[1][length]++; //记录a数组中长度为length的循环节个数        }        for(int i=0; i<m; i++)            scanf("%d",&b[i]);        for(int i=0; i<m; i++)        {            ll k=i;            ll length=0;            while(b[k]!=-1)            {                length++;                int t=k;                k=b[k];                b[t]=-1;            }            if(length!=0)                circle[2][length]++; //记录b数组中长度为length的循环节个数        }        long long ans=1;        for(int i=1; i<=n; i++)        {            if(circle[1][i]!=0)            {                int lim=(int)sqrt(i+0.5),sum=0;//稍微优化                for(int j=1; j<=lim; j++)                {                    if(i%j==0) //a的循环节长度能够整除b的循环节长度                    {                        sum=sum+circle[2][j]%Mod*j%Mod;                        //a的其中一个循环节在b中的各个因子计算:(长度为w)循环节个数*(w)循环节长度,再取和                        sum=sum%Mod;                        if(j*j!=i)                        {                            sum=sum+circle[2][i/j]%Mod*(i/j)%Mod;                            sum=sum%Mod;                        }                    }                }                for(int j=1; j<=circle[1][i]; j++)                {                    ans=ans*sum; //将所有sum相乘得结果                    ans=ans%Mod;                }            }        }        cout<<"Case #"<<ca++<<": "<<ans<<endl;    }}