HDU 6038 函数问题

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本文探讨了一个复杂的函数计数问题，给定两个排列序列a和b，寻找满足特定条件的函数数量。通过对序列循环特性的深入分析，提出了一种有效的解决策略，并通过代码实现了算法。

Function

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1649 Accepted Submission(s): 773

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

(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 #

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

Sample Output

Case #1: 4
Case #2: 4

题意:

给你一个a序列，代表0到n-1的排列；一个b序列代表0到m-1的排列。问你可以找出多少种函数关系，满足f(i)=b[f(a[i])];
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<stdio.h>
#include<string.h>
#include<vector>
using namespace std;
#define ll long long
#define mod 1000000007
#define N 100050

vector<int>vec;
vector<int>F[N];

void init(){
	for(int i = 1;i <= N - 10;i++){
		for(int j = i;j <= N - 10;j += i){
			F[j].push_back(i);
		}
	}
}

int a[N],b[N],vis[N],blen[N];

int dfs(int x,int *p){
	if(vis[x])
		return 0;
	vis[x] = 1;
	return dfs(p[x],p) + 1;
}

int main(){
	int n,m,Case = 1;
	init();
	while(scanf("%d %d",&n,&m) != EOF){
		vec.clear();
		for(int i = 0;i < n;i++){
			scanf("%d",&a[i]);
		}
		for(int i = 0;i < m;i++){
			scanf("%d",&b[i]);
		}
		memset(vis,0,sizeof(vis));
		for(int i = 0;i < n;i++){
			if(!vis[i]){
				vec.push_back(dfs(i,a));
			}
		}
		memset(vis,0,sizeof(vis));
		memset(blen,0,sizeof(blen));
		for(int i = 0;i < m;i++){
			if(!vis[i]){
				blen[dfs(i,b)]++;
			}
		}
		ll ans = 1;
		for(int i = 0;i < vec.size();i++){
			ll temp = 0;
			for(int j = 0;j < F[vec[i]].size();j++){
				temp += F[vec[i]][j] * blen[F[vec[i]][j]];
				temp %= mod;
			}
			ans *= temp;
			ans %= mod;
		}
		printf("Case #%d: %lld\n",Case++,ans);
	}
}