[USACOOpen2020P]Exercise

English (en)
Farmer John has come up with a new morning exercise routine for the cows (again)!
As before, Farmer John's N cows (1≤N≤7500) are standing in a line. The i-th cow from the left has label i for each 1≤i≤N. He tells them to repeat the following step until the cows are in the same order as when they started.

Given a permutation A of length N, the cows change their order such that the i-th cow from the left before the change is Ai-th from the left after the change.

For example, if A=(1,2,3,4,5) then the cows perform one step and immediately return to the same order. If A=(2,3,1,5,4), then the cows perform six steps before returning to the original order. The order of the cows from left to right after each step is as follows:

0 steps: (1,2,3,4,5)
1 step: (3,1,2,5,4)
2 steps: (2,3,1,4,5)
3 steps: (1,2,3,5,4)
4 steps: (3,1,2,4,5)
5 steps: (2,3,1,5,4)
6 steps: (1,2,3,4,5)

Compute the product of the numbers of steps needed over all N! possible permutations A of length N.

As this number may be very large, output the answer modulo M (10⁸≤M≤10⁹+7, M is prime).

Contestants using C++ may find the following code from KACTL helpful. Known as the Barrett reduction, it allows you to compute a%b several times faster than usual, where b>1 is constant but not known at compile time. (we are not aware of such an optimization for Java, unfortunately).

#include <bits/stdc++.h>
using namespace std;

typedef unsigned long long ull;
typedef __uint128_t L;
struct FastMod {
ull b, m;
FastMod(ull b) : b(b), m(ull((L(1) << 64) / b)) {}
ull reduce(ull a) {
ull q = (ull)((L(m) * a) >> 64);
ull r = a - q * b; // can be proven that 0 <= r < 2*b
return r >= b ? r - b : r;
}
};
FastMod F(2);

int main() {
int M = 1000000007; F = FastMod(M);
ull x = 10ULL*M+3;
cout << x << " " << F.reduce(x) << "\n"; // 10000000073 3
}

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