【HDU 6038 Problem Description】+ 思维

Function

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 365 Accepted Submission(s): 142

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

题解 ：
考虑置换 aa 的一个循环节，长度为 ll ，那么有 f(i)=bf(ai)=bbf(aai)=b⋯bf(i)l times b 。
那么 f(i)f(i) 的值在置换 bb 中所在的循环节的长度必须为 ll 的因数。
而如果 f(i)f(i) 的值确定下来了，这个循环节的另外 l - 1l−1 个数的函数值也都确定下来了。
答案就是 \
sum_{i = 1}^{k} \sum_{j | l_i} {j \cdot cal_j}∑​i=1​k∑​j∣l​ij⋅cal​j
​改为 \
prod_{i = 1}^{k} \sum_{j | l_i} {j \cdot cal_j}∏​i=1​k​​ ∑​j∣l​i j⋅cal​j
​，其中 kk 是置换 aa 中循环节的个数， l_il​i
​​ 表示置换 aa 中第 ii 个循环节的长度， cal_jcal​j
​​ 表示置换 bb 中长度为 jj 的循环节的个数。
时间复杂度是 O(n+m) 。

AC代码：

#include<cstdio>#include<cstring>using namespace std;typedef long long LL;const int MAX = 1e5 + 10;const int mod = 1e9 + 7;int a[MAX],b[MAX],num[2][MAX],vis[MAX],n,m;void dfs(int *p,int o,int nl,int x){    if(vis[o]){        num[x][nl]++; // 循环节为 nl 的个数        return ;    }    vis[o] = 1;    dfs(p,p[o],nl + 1,x);}int main(){    int nl = 0;    while(~scanf("%d %d",&n,&m)){        memset(num,0,sizeof(num));        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]);        for(int i = 0; i < n; i++)            if(!vis[i]) // 计算 a 里面的循环体                dfs(a,i,0,0);        memset(vis,0,sizeof(vis));        for(int i = 0; i < m; i++)            if(!vis[i]) // 计算 b 里面的循环体                dfs(b,i,0,1);        LL ans = 1;        for(int i = 1; i <= n; i++)            if(num[0][i]){                LL sum = 0;                for(int j = 1; j * j <= i; j++){                    if(i % j == 0){                        sum = (sum + (LL)j * num[1][j] % mod) % mod;                        if(j * j != i)                            sum = (sum + (LL)i / j * num[1][i / j] % mod) % mod;                    }                 }                for(int j = 1; j <= num[0][i]; j++)                    ans = ans * sum % mod;            }        printf("Case #%d: %lld\n",++nl,ans);    }    return 0;}