HDU 6038 函数问题

Function

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1649    Accepted Submission(s): 773

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

 

题意:

给你一个a序列，代表0到n-1的排列；一个b序列代表0到m-1的排列。问你可以找出多少种函数关系，满足f(i)=b[f(a[i])]; 
分析：这个主要是找循环节 
比如说：如果 a 序列是 2 0 1 那么我们可以发现

f(0) = b[f(a[0])] = b[f(2)] 
f[1] = b[f(a[1])] = b[f(0)] 
f[2] = b[f(a[2])] = b[f(1)]

那么f(0) f(1) f(2) 也是循环的 如果想找出这样的函数，必须值域里也存在同样长度的循环节或者存在其约数长度的循环节。 
那么就是找两个序列的循环节，对于定义域里面的每一个循环节都找出来有多少种和他对应的。最后乘起来就是答案了

#include<stdio.h>#include<string.h>#include<vector>using namespace std;#define ll long long#define mod 1000000007#define N 100050vector<int>vec;vector<int>F[N];void init(){for(int i = 1;i <= N - 10;i++){for(int j = i;j <= N - 10;j += i){F[j].push_back(i);}}}int a[N],b[N],vis[N],blen[N];int dfs(int x,int *p){if(vis[x])return 0;vis[x] = 1;return dfs(p[x],p) + 1;}int main(){int n,m,Case = 1;init();while(scanf("%d %d",&n,&m) != EOF){vec.clear();for(int i = 0;i < n;i++){scanf("%d",&a[i]);}for(int i = 0;i < m;i++){scanf("%d",&b[i]);}memset(vis,0,sizeof(vis));for(int i = 0;i < n;i++){if(!vis[i]){vec.push_back(dfs(i,a));}}memset(vis,0,sizeof(vis));memset(blen,0,sizeof(blen));for(int i = 0;i < m;i++){if(!vis[i]){blen[dfs(i,b)]++;}}ll ans = 1;for(int i = 0;i < vec.size();i++){ll temp = 0;for(int j = 0;j < F[vec[i]].size();j++){temp += F[vec[i]][j] * blen[F[vec[i]][j]];temp %= mod;}ans *= temp;ans %= mod;}printf("Case #%d: %lld\n",Case++,ans);}}