문제
Farmer John has come up with a new morning exercise routine for the cows (again)!
As before, Farmer John's cows () are standing in a line. The -th cow from the left has label for each . He tells them to repeat the following step until the cows are in the same order as when they started.
Given a permutation of length , the cows change their order such that the -th cow from the left before the change is -th from the left after the change.
For example, if then the cows perform one step and immediately return to the same order. If , 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 step: 2 steps: 3 steps: 4 steps: 5 steps: 6 steps:
Compute the product of the numbers of steps needed over all possible permutations of length .
As this number may be very large, output the answer modulo (, is prime).
Contestants using C++ may find the following code from KACTL helpful. Known as the Barrett reduction, it allows you to compute several times faster than usual, where 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 }
입력
The first line contains and .
출력
A single integer.
예제 입력 1
5 1000000007
예제 출력 1
369329541
점수
Test case 2 satisfies .Test cases 3-5 satisfy .Test cases 6-8 satisfy .Test cases 9-12 satisfy .Test cases 13-16 satisfy no additional constraints.
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