| 시간 제한 | 메모리 제한 | 제출 | 정답 | 맞힌 사람 | 정답 비율 |
|---|---|---|---|---|---|
| 2 초 | 512 MB | 54 | 18 | 16 | 37.209% |
Farmer John has come up with a new morning exercise routine for the cows (again)!
As before, Farmer John's $N$ cows (1ドル\le N\le 7500$) are standing in a line. The $i$-th cow from the left has label $i$ for each 1ドル\le i\le N$. He tells them to repeat the following step until the cows are in the same order as when they started.
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),ドル 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:
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 $M$ (10ドル^8\le M\le 10^9+7,ドル $M$ 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
}
The first line contains $N$ and $M$.
A single integer.
5 1000000007
369329541
For each 1ドル\le i\le N,ドル the $i$-th element of the following array is the number of permutations that cause the cows to take $i$ steps: $[1,25,20,30,24,20].$ The answer is 1ドル^1\cdot 2^{25}\cdot 3^{20}\cdot 4^{30}\cdot 5^{24}\cdot 6^{20}\equiv 369329541\pmod{10^9+7}$.