Function
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1557 Accepted Submission(s): 738
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
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int inf = 1e9;
const int mod = 1e9+7;
int book[111111];
int ans[111111];
int ans2[111111];
ll quick_pow(ll a,ll b){
ll ans = 1;
while(b){
if(b&1) ans = (ans*a) % mod;
b >>= 1;
a = (a*a) % mod;
}
return ans;
}
int main(){
std::ios::sync_with_stdio(false);
int n,m;
int caset = 0;
while(cin>>n>>m){
int ansc[111111] = {0};
int cnt = 0;
for(int i=0; i<n; i++)
cin>>book[i];
int mapp[111111] = {0};
for(int i=0; i<n; i++)
if(!mapp[i]){
int count = 0;
int tmp = i;
do{
mapp[tmp] = 1;
tmp = book[tmp];
count++;
}while(tmp != i);
int flag = 0;
for(int j=0; j<cnt; j++){
if(ans[j] == count)
ansc[j]++,flag = 1;
}
if(!flag){
ans[cnt++] = count;
ansc[cnt-1]++;
}
}
int ans2c[111111] = {0};
int cnt2 = 0;
for(int i=0; i<m; i++)
cin>>book[i];
memset(mapp, 0, sizeof(mapp));
for(int i=0; i<m; i++)
if(!mapp[i]){
int count = 0;
int tmp = i;
do{
mapp[tmp] = 1;
tmp = book[tmp];
count++;
}while(tmp != i);
int flag = 0;
for(int j=0; j<cnt2; j++){
if(ans2[j] == count)
ans2c[j]++,flag = 1;
}
if(!flag){
ans2[cnt2++] = count;
ans2c[cnt2-1]++;
}
}
ll anss = 1;
for(int i=0; i<cnt; i++){
ll tmp = 0;
for(int j=0; j<cnt2; j++)
if(ans[i]%ans2[j] == 0) tmp += ans2c[j]*ans2[j] % mod;
anss =( anss * quick_pow(tmp, ansc[i]) ) %mod;
}
cout<<"Case #"<<++caset<<": ";
cout<<anss<<endl;
}
}