[游戏开发]Codeforces:1658B Marin and Anti-coprime Permutation题解

Marin wants you to count number of permutations that are beautiful. A beautiful permutation of length n is a permutation that has the following property:
gcd(1?p1,2?p2,…,n?pn)>1,
where gcd is the greatest common divisor.

A permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).

Input
The first line contains one integer t (1≤t≤103) — the number of test cases.

Each test case consists of one line containing one integer n (1≤n≤103).

Output
For each test case, print one integer — number of beautiful permutations. Because the answer can be very big, please print the answer modulo 998244353.

分析：当n为奇数时，permutations的数量为0
当n为偶数时，permutations的数量为pow((n/2)!,2)
感叹号为阶层

		例子：n=6    A(3,3)*A(3,3)=36
		即排列组合n/2的偶数位3个与n/2的奇数位3个排列组合

原因：当n为偶数时gcd(1 * p1, 2 * p2…n * pn)
p1，p3，p5…只能放偶数
p2，p4，p6…只能放奇数
此时 p11,p22,p3*3 全为偶数
最大公约数为2

import java.util.Scanner;


public class Main {
    public static void main(String[] args)  {
        Scanner in=new Scanner(System.in);
        int t=in.nextInt();
        for (int i=0;i<t;i++)
        {
            int n=in.nextInt();
            if ((n&1)==1)//为奇数
            {
                System.out.println(0);
            }
            else
            {
            	//n除以2
                n=n/2;
                long sum=1l;
                for (int j=2;j<=n;j++)//n/2
                {
                    sum=sum*j*j%998244353;
                }
                System.out.println(sum);
            }
        }

    }
}

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