hdu 6038

Function

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1353    Accepted Submission(s): 623

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 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 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

题意：给出两个排列，a和b，求满足映射的f[i]=b_f[a[i]]的个数。

其实相当于排列b也是一个映射，把两个映射转化成两个图。

因为是1-n的排列，所以根据映射关系组成的一定是一个一个的环。

为了将f的映射和b的对应起来，对于f中的每个环，当b中的环的长度为f中的环的因数时，则可以匹配。最后再组合一下答案即可。

ac代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<sstream>
#include<vector>
#include<stack>
#include<queue>
#include<cmath>
#include<map>
#include<set>
using namespace std;
typedef long long ll;
typedef pair<int,int> pll;
const int INF = 0x3f3f3f3f;
const int maxn = 1e5 + 5;
const int mod = 1e9+7;


int n, m;


int a[maxn], b[maxn];
int vis[maxn];


vector<int> A;
vector<int> B;


int main()
{
    //freopen("in.txt","r",stdin);
    int kase=0;
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=0;i<n;i++)  scanf("%d",&a[i]);
        for(int i=0;i<m;i++)  scanf("%d",&b[i]);


        A.clear();
        memset(vis,0,sizeof(vis));
        for(int i=0;i<n;i++)
        {
            int cur=i,cnt=1;
            if(!vis[i])
            {
                vis[i]=1;
                while(a[cur]!=i)
                {
                    cur=a[cur];
                    vis[cur]=1;
                    cnt++;
                }
                A.push_back(cnt);
            }
        }


        B.clear();
        memset(vis,0,sizeof(vis));
        for(int i=0;i<m;i++)
        {
            int cur=i, cnt=1;
            if(!vis[i])
            {
                vis[i]=1;
                while(b[cur]!=i)
                {
                    cur=b[cur];
                    vis[cur]=1;
                    cnt++;
                }
                B.push_back(cnt);
            }
        }


        ll ans=1;
        for(int i=0;i<A.size();i++)
        {
            ll tmp=0;
            for(int j=0;j<B.size();j++)
            {
                if(A[i]%B[j]==0)  tmp=(tmp+B[j])%mod;
            }
            ans=ans*tmp%mod;
        }
        printf("Case #%d: %d\n",++kase,ans);
    }
    return 0;
}