Function

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 838 Accepted Submission(s): 369

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

比赛的时候想到了环这个东西，怀疑自己的思路，没有继续向下推了。果然专注度和练习的强度不够。继续加油。

AC代码：

#include <iostream>

#include <cstdio>

#include <cstring>

#include <queue>

#include <stack>

#include <vector>

using namespace std;

typedef long long ll;

ll a[],b[];

vector<ll> edgea,edgeb;

const ll mod=1e9+;

ll n,m;

void solve(int Case)

{

    ll visa[],visb[];

    memset(visb,,sizeof(visb));

    memset(visa,,sizeof(visa));

    for(int i=;i<m;i++)

    {

        if(visb[i]) continue;

        ll ret=;

        ll next_pos=b[i];

        visb[next_pos]=;

        while(next_pos!=i)

        {

            ret++;

            next_pos=b[next_pos];

            visb[next_pos]=;

        }

       // cout<<ret<<endl;

        edgeb.push_back(ret);

    }

    ll f=;

    for(int i=;i<n;i++)

    {

        if(visa[i]) continue;

        ll ret=;

        ll next_pos=a[i];

        visa[next_pos]=;

        while(next_pos!=i)

        {

            ret++;

            next_pos=a[next_pos];

            visa[next_pos]=;

        }

        ll temp=;

        //cout<<ret<<endl;

        for(int j=;j<edgeb.size();j++)

        {

            if(ret%edgeb[j]== || ret==edgeb[j]) temp=(temp+edgeb[j])%mod;

        }

        f=f*temp%mod;

    }

    printf("Case #%d: %d\n",Case,f);

}

int main()

{

    cin.sync_with_stdio(false);

    int Case=;

    while(cin>>n>>m)

    {

        edgea.clear();

        edgeb.clear();

        for(int i=;i<n;i++) cin>>a[i];

        for(int j=;j<m;j++) cin>>b[j];

        solve(Case++);

    }

    return ;

}

巴特西

2017多校赛 Function

Function

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