2017 Multi-University Training Contest

Function

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 842    Accepted Submission(s): 370

Problem Description

You are given a permutationa 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 from0 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

 

题目大意：

给你一个数组A，和一个数组B，数组A是【0~n-1】的排咧，数组B是【0~m-1】的排列。

现在定义F(i)=bF(ai)；问有多少种取值，使得F(i)全部合法

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<cstdio>#include<map>#include<cstring>#define mo 1000000007using namespace std;typedef long long ll;const int maxn = 1e5 + 5;ll a[maxn];ll b[maxn];ll num1[maxn], num2[maxn];int main(){    int n, m, kase = 0;    while (~scanf("%d%d", &n, &m))    {        memset(num2, 0, sizeof(num2));        int totfa = 0;        for (int i = 0; i < n; i++)        {            scanf("%lld", &a[i]);        }        for (int i = 0; i < n; i++)        {            ll tot = 0;            int k = i;            while (a[k] != -1)            {                tot++;                int t = k;                k = a[k];                a[t] = -1;            }            if (tot)            {                num1[++totfa] = tot;            }        }        for (int j = 0; j < m; j++)        {            scanf("%lld", &b[j]);        }        for (int i = 0; i < m; i++)        {            ll tot = 0;            int k = i;            while (b[k] != -1)            {                tot++;                int t = k;                k = b[k];                b[t] = -1;            }            if (tot)            {                num2[tot]++;            }        }        ll ans = 1;        for (int i = 1; i <= totfa; i++)        {            ll ansl = 0;            for (int j = 1; j * j <= num1[i]; j++)            {                if (num1[i] % j == 0)                {                    if (j * j == num1[i])                    {                        ansl += num2[j] * j;                    }                    else                    {                        ansl += num2[j] * j + num2[num1[i] / j] * num1[i] / j;                    }                }            }            ans = (ans * ansl) % mo;        }        printf("Case #%d: %lld\n", ++kase, ans);    }}