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.
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.
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
启发博客：http://blog.csdn.net/changjiale110/article/details/76448385
以下题解摘自此博客
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。
寻找都推关系，根据已知的a数组和b数组，找出a，b两个数组于f的对应关系。
因为提上给出的f(i)=b[f(a[i])]，此时我们可以转换一下，将a[i]看作要求的参数，则上面的式子可以转 换为f(i)=b[f(a[i])]=b[b[f(a[a[i]])]]，如果我们已知这样转换下去，直到将最后的结果能够转换成f(i)为止 的话。可发现，最里面的一成是i->a[i]->a[a[i]]···-.i的一个环，这个环的循环可以决定f()函数的循环， 进而决定外面的b的循环，相当于f的循环可以由a和b的环来决定。
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。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<queue> 5 #include<algorithm> 6 #include<cmath> 7 #include<vector> 8 using namespace std; 9 int n,m; 10 const int maxn=100005; 11 const int mod=1e9+7; 12 int a[maxn],b[maxn],lenb[maxn]; 13 //lenb[i]记录的是b中循环长度为i的环个数 14 vector<int>aa;//构建环 15 vector<int>fac[maxn];//记录因子 16 bool vis[maxn]; 17 //找环，并返回环的大小 18 int dfs(int i,int*c) 19 { 20 if(vis[i]) 21 return 0; 22 vis[i]=1; 23 return dfs(c[i],c)+1; 24 } 25 26 void get_fac() 27 { 28 for(int i=1;i<=100000;i++) 29 for(int j=i;j<=100000;j+=i) 30 fac[j].push_back(i); 31 //fac[j]里面保存的是长度为j的环的因子 32 } 33 34 int main() 35 { 36 int T=1; 37 get_fac(); 38 while(~scanf("%d%d",&n,&m)) 39 { 40 for(int i=0;i<n;i++) 41 scanf("%d",&a[i]); 42 for(int i=0;i<m;i++) 43 scanf("%d",&b[i]); 44 aa.clear(); 45 memset(vis,false,sizeof(vis)); 46 for(int i=0;i<n;i++) 47 { 48 if(vis[i])continue; 49 aa.push_back(dfs(i,a));//aa数组中存下a中每个环的长度 50 } 51 memset(vis,false,sizeof(vis)); 52 memset(lenb,0,sizeof(lenb)); 53 for(int i=0;i<m;i++) 54 { 55 if(vis[i]) 56 continue; 57 lenb[dfs(i,b)]++; 58 } 59 long long ans=1; 60 int L=aa.size(); 61 //根据a的每个环去找b约数环 62 for(int i=0;i<L;i++) 63 { 64 int lena=aa[i],ll=fac[lena].size(); 65 long long res=0; 66 for(int j=0;j<ll;j++) 67 { 68 int lb=fac[lena][j];//lb是长度为lena的环的一个因子 69 res=(res+(long long)lb*lenb[lb])%mod;//乘上长度为这个因子的环的个数 70 } 71 ans=(ans*res)%mod; 72 } 73 printf("Case #%d: %lld\n",T++,ans); 74 } 75 return 0; 76 }
