HDU 6038 Function

Function

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 607    Accepted Submission(s): 243

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

从集合b中选择n个数构成f[i]，问符合f[i] = b[ f[ a[i] ]  ]的映射关系有多少种。

第一个样例  a={1,0,2}   b={0,1}

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）

那么要使得满足条件带入的b必须成环而且是 a环长度的因子

按照这个思路就可以得到答案了~

很套路的题。

#include <bits/stdc++.h>using namespace std;const int MAXN = 1e5+7;const long long mod = 1e9+7;int n,m;bool vis[MAXN];int a[MAXN],b[MAXN];vector<int>circle_a,factor[MAXN];//寻找环int dfs(int u,int *p){    if(vis[u])return 0;    vis[u] = 1;    return dfs(p[u],p) + 1;}//记录长度为下标的环的个数int lenb[MAXN];//得到因子void get_factor(){    for(int i = 1; i <= 100000; ++i)    {        for(int j = i; j <= 100000; j += i)factor[j].push_back(i);    }}int main(){    get_factor();    int ca = 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]);        circle_a.clear();        memset(vis,0,sizeof vis);        for(int i = 0; i < n; ++i)        {            if(vis[i])continue;            circle_a.push_back(dfs(i,a));        }        memset(vis,0,sizeof vis);        memset(lenb,0,sizeof lenb);        for(int i = 0; i < m; ++i)        {            if(vis[i])continue;            lenb[dfs(i,b)]++;;        }        long long ans = 1;        for(int i = 0,l = circle_a.size(); i < l; ++i)        {            int la = circle_a[i];            long long res = 0;            for(int j = 0,l1 = factor[la].size(); j < l1; ++j)            {                int lb = factor[la][j];                res = (res + (long long)lb*lenb[lb])%mod;            }            ans = ans*res%mod;        }        printf("Case #%d: %I64d\n",++ca,ans);    }    return 0;}