6038 Function

Function

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1300    Accepted Submission(s): 599

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

 

Source

2017 Multi-University Training Contest - Team 1 

思路：

函数的映射。每个f（x）与ax关联关系链成环。必须有相应的b与a对应（一个a只能有一个b，一个b可以有多个a对应）。

先处理a和b有几个环  ，再 b组成（每个）a环的种类数。  一个有k个元素的b环有k种可能有num[k]=x个有k个元素的b 环就是有 num[k]*k种可能。a环元素有k个元素同时可以由k的因子成环（如果k为4 ，k可以由4个元素的、2个元素的、1个元素的环组成）所以对于一个a环可以由sum=num[x1]*x1+num[x2]*x2+num[x3]*x3…. （x1,x2,x3为k的因子）。  如果有t个有k个因子的a环 ans*=sum1^t1；所有a环答案相乘就是最后的答案。

AC：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <cmath>typedef long long LL;using namespace std;int vis[100100]={0};int mem[100100]={0},VIS[100100]={0};int cira[100100]={0},cirb[100100]={0};int main(){    int n,m;    int v=1;    while(scanf("%d%d",&n,&m)!=EOF)    {        int a[100100]={0},b[100100]={0};        for(int i=0;i<n;i++)scanf("%d",&a[i]);        int s,next,num=1;        memset(VIS,0,sizeof(VIS));        memset(cira,0,sizeof(cira));        memset(cirb,0,sizeof(cirb));        memset(vis,0,sizeof(vis));        memset(mem,0,sizeof(mem));        ///the circle of A        int d=0;        for(int i=0;i<n;i++)        {            s=i,next=a[i],num=1;            if(VIS[s]==1)continue;            VIS[s]=1;            //printf("s= %d \n",s);            while(next!= s)            {                VIS[next]=1;                next=a[next];                num++;            }            cira[num]++;            if(!vis[num])mem[d++]=num;            vis[num]=1;        }        //for(int i=0;i<10;i++)printf("ciara[%d]=%d ",i,cira[i]);        for(int i=0;i<m;i++)scanf("%d",&b[i]);        memset(VIS,0,sizeof(VIS));        ///the circle of B        for(int i=0;i<m;i++)        {            s=i,next=b[i],num=1;            if(VIS[s]==1)continue;            VIS[s]=1;            //printf("s= %d \n",s);            while(next!= s)            {                VIS[next]=1;                next=b[next];                num++;            }            cirb[num]++;        }        //for(int i=0;i<10;i++)printf("ciarb[%d]=%d ",i,cirb[i]);        LL ans=1;        //cout<<"a:";        for(int i=0;i<d;i++)        {            int k=mem[i];            //printf("%d \n",k);            LL temp=0;            for(int j=1;j<=k;j++)            {                if(k%j==0)temp+=j*cirb[j];            }            //cout<<"temp:"<<temp<<endl;            LL T=1;            for(int j=1;j<=cira[k];j++)                T*=temp;            ans*=T;        }        //cout<<endl;        printf("Case #%d: %lld\n",v++,ans);    }}