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 2 1 0 2 0 1 3 4 2 0 1 0 2 3 1
Sample Output Case #1: 4 Case #2: 4 题目大意：有两个，数组a是[0~n-1]的排列，数组b是[0~m-1]的排列。现在定义f(i)=b[f(a[i])]； 问f(i)有多少种取值，使得表达式f(i)=b[f(a[i])]全部合法。 解题思路：猛的一看可能感觉无从下手，不知道该怎么办，但由题目f(i)=b[f(a[i])]可以递推出f(i)=b[f(a[i])]=b[b[f(a[a[i]])]]，以此类推，可以一直递推下去，我们可以得到i->a[i]->a[a[i]]->a[a[a[i]]]··· ···->i这样的一个环以第一个样例 a={1,0,2} b={0,1}为例： 那么f(0)=b[f(1)] f(1)=b[f(0)] f(2)=b[f(2)] 这里有两个环分别为 f(0)->f(1) 和f（2） 所以我们的任务就是在b中找环，该环的长度必须为a中环的长度的约数。为什么必须的是约数呢？因为如果b的环的长度是a的环的长度的约数的话，那也就意味着用b这个环也能构成a这个环，只不过是多循环了几次而已。然后找到a中所有环的方案数，累乘便是答案。 为什么要累乘呢？我最开始一直以为要累加。这个就用到了排列组合的思想，因为肯定要f(i)肯定要满足所有的数，而a中的每个环都相当于从a中取出几个数的方案数，所以总共的方案数应该累乘。
#include <bits/stdc++.h> const int mod=1e9+7; typedef long long LL ; using namespace std; int a[100005],b[100005],vis[100005]; vector<int> fac[100010]; void get_fac() { for(int i = 1; i <= 100000; i++){ for(int j = i; j <= 100000; j+=i) fac[j].push_back(i); } } int dfs(int aa[],int n) { if(vis[n]) return 0; vis[n] = 1; return dfs(aa,aa[n])+1; } int main() { int n,m,c=1; get_fac(); while(~scanf("%d %d",&n,&m)){ memset(vis,0,sizeof(vis)); for(int i = 0; i < n; i++) scanf("%d",&a[i]); for(int i = 0; i < m; i++) scanf("%d",&b[i]); vector<int> A; int B[100005]; memset(B,0,sizeof(B)); for(int i = 0; i < n; i++) if(!vis[i]) A.push_back(dfs(a,i)); memset(vis,0,sizeof(vis)); for(int i = 0; i < m; i++) if(!vis[i]) B[dfs(b,i)]++; int la,lb,len; LL ans = 1; for(int i = 0; i < A.size(); i++) { la = A[i]; LL temp = 0; len = fac[la].size(); for(int j = 0; j < len; j++){ lb = fac[la][j]; temp = (temp + (long long)(lb*B[lb]))%mod; } ans = (ans*temp)%mod; } printf("Case #%d: %lld\n",c++,ans); } return 0; }