学长讲座讲过的，代码也讲过了，然而，当时上课没来听，听代码的时候也一脸o((⊙﹏⊙))o
我的妈呀，语文不好是硬伤，看题意看了好久好久好久(死一死)。。。
数学+思维题，代码懂了，也能写出来，但是还是有一点不懂，明天继续。
感谢我的队友不嫌弃我是他的猪队友(可能他心里已经骂了无数次我是猪队友了_(:з」∠)_ )
Function
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2021 Accepted Submission(s): 956
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.
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.
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
Source
题意:吐槽啊，看不懂题意，后来推出来了，就是满足f(i)=bf(ai)这个式子。
函数套函数，假设a这个数组有5个数，那么就是f(0),f(1),f(2),f(3),f(4)这5个。
然后就是推。。。
对于f(0)，i为0，i为a的下标，那么就是ai为a0，就是f(0)==bf(a0),然后看看a0对应的值是多少，这个值就是b的下标。就是这个意思。
然后就是，f这个函数值域是b数组中的任意一个数。假设a0对应的值是3，那么就是f(0)==bf(3)，
f(3)的值是多少呢？就从b的数组中任选一个，如果这个数能够依次往下推，推到和最初的值相等就是符合条件的。推的这个过程就是一个循环节。
通过总结就可以发现，i和a数组有关，i和b数组也有关(和b有关的i不是和a有关的i)，通过分别找出来a和b的循环节，就可以把个数相乘就是结果。
但是发现找循环节的时候，必须是b的循环节的长度是a的循环节的长度的约数才可以(就是b的长度这个数可以被a的长度的数整除)。
但是我这里还不太懂为什么可以，队友给我讲的，明天继续研究，先写完题解滚去补作业。。。
代码直接看我队友的代码吧，我也是看的他的写的。传送门:_(:з」∠)_
然后贴一下余学长的代码_(:з」∠)_ (别打我。。。)
代码:
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<string> 5 #include<queue> 6 #include<algorithm> 7 #include<stack> 8 #include<cstring> 9 #include<vector> 10 #include<list> 11 #include<bitset> 12 #include<set> 13 #include<map> 14 #include<time.h> 15 using namespace std; 16 typedef long long ll; 17 const int maxn=100000+10; 18 const int mod=1e9+7; 19 map<int,int>mp1,mp2; 20 map<int,int>::iterator it1,it2; 21 int a[maxn],b[maxn],vis[maxn]; 22 int n,m; 23 void solve(int n,int *f,map<int,int> &mp) 24 { 25 memset(vis,0,sizeof(vis)); 26 int j,k; 27 for(int i=0;i<n;i++) 28 { 29 j=i,k=0; 30 while(!vis[j]) 31 { 32 k++; 33 vis[j]=1; 34 j=f[j]; 35 } 36 if(k) 37 mp[k]++; 38 } 39 } 40 int main() 41 { 42 int kase=0; 43 while(scanf("%d%d",&n,&m)!=EOF) 44 { 45 for(int i=0;i<n;i++) 46 scanf("%d",&a[i]); 47 for(int i=0;i<m;i++) 48 scanf("%d",&b[i]); 49 mp1.clear(); 50 mp2.clear(); 51 solve(n,a,mp1); 52 solve(m,b,mp2); 53 ll ans=1; 54 for(it1=mp1.begin();it1!=mp1.end();it1++) 55 { 56 ll cnt=0,x=it1->first,t=it1->second; 57 for(it2=mp2.begin();it2!=mp2.end();it2++) 58 { 59 int y=it2->first,num=it2->second; 60 if(x%y==0)cnt=(cnt+y*num)%mod; 61 } 62 for(int i=1;i<=t;i++) 63 ans=(ans*cnt)%mod; 64 } 65 printf("Case #%d: %lld ",++kase,ans); 66 } 67 return 0; 68 }
学长的代码比我队友的快_(:з」∠)_
溜了溜了。不想喝拿铁咖啡。
加油加油_(:з」∠)_