hud 6038 Function

最新推荐文章于 2024-09-07 17:13:23 发布

Merry_hj

最新推荐文章于 2024-09-07 17:13:23 发布

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本文探讨了一个关于函数f的问题，给定两个排列数组a和b，寻找满足特定条件的函数f的数量。通过分析循环结构并利用数学方法，文章提供了一种高效求解策略。

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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;
}