Function HDU

[题目链接](http://acm.hdu.edu.cn/showproblem.php?pid=6038)

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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 1e9+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 test case, output " Case #x: y" in one line (without quotes), where x indicates the case number starting from 1 and ydenotes 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

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题意:

给出两个序列 a 和 b ，求满足f(i)=bf(ai) 的函数个数。

思路:

根据样例 1 我们可以得到：

f(0)=bf(1)

f(1)=bf(0)

f(2)=bf(2)

官方题解很清楚

代码:

#include<bits/stdc++.h>using namespace std;typedef long long LL;const int mod = 1e9+7;const int maxn = 1e5 + 10;int a[maxn],b[maxn];bool vis[maxn];vector<int> q1,q2;int main(){    int n, m;    int kase = 0;    while(scanf("%d %d", &n, &m) != EOF){        for(int i = 0; i < n; ++i)            scanf("%d", a+i);        for(int i = 0; i < m ;++i)            scanf("%d", b+i);        printf("Case #%d: ", ++kase);        memset(vis,false,sizeof(vis));        q1.clear();        q2.clear();        for(int i = 0; i < n; ++i){            int cur = i;            int cnt = 1;            if(vis[i]) continue;            while(a[cur] != i){                cur =a[cur];                vis[cur] = true;                ++cnt;            }q1.push_back(cnt);        }memset(vis,false,sizeof(vis));         for(int i = 0; i < m; ++i){            int cur = i;            int cnt = 1;            if(vis[i]) continue;            while(b[cur] != i){                cur =b[cur];                vis[cur] = true;                ++cnt;            }q2.push_back(cnt);        }int ans = 1;        int len1 = q1.size();        int len2 = q2.size();        for(int i = 0; i < len1; ++i){            int res = 0;            for(int j = 0; j < len2; ++j)                if(q1[i] % q2[j] == 0)                    res = (res + q2[j]) % mod;            ans = ans * res % mod;        }printf("%d\n",ans);    }return 0;}