F. Zero Remainder

F. Zero Remainder Sum

You are given a matrix a of size n×m consisting of integers.

You can choose no more than ⌊m2⌋ elements in each row. Your task is to choose these elements in such a way that their sum is divisible by k and this sum is the maximum.

In other words, you can choose no more than a half (rounded down) of elements in each row, you have to find the maximum sum of these elements divisible by k.

Note that you can choose zero elements (and the sum of such set is 0).

Input
The first line of the input contains three integers n, m and k (1≤n,m,k≤70) — the number of rows in the matrix, the number of columns in the matrix and the value of k. The next n lines contain m elements each, where the j-th element of the i-th row is ai,j (1≤ai,j≤70).

Output
Print one integer — the maximum sum divisible by k you can obtain.

Examples
inputCopy
3 4 3
1 2 3 4
5 2 2 2
7 1 1 4
output
24
input
5 5 4
1 2 4 2 1
3 5 1 2 4
1 5 7 1 2
3 8 7 1 2
8 4 7 1 6
output
56
Note
In the first example, the optimal answer is 2 and 4 in the first row, 5 and 2 in the second row and 7 and 4 in the third row. The total sum is 2+4+5+2+7+4=24.

题意：
给你一个整数n，代表n个人，每个人都不一样；
你的任务是找出n个人可以跳两个回合舞的方法，
每一个回合舞都是由n/2个人组成的。

输出所有方法数。

看完题目，看样例，直觉告诉我，可以直接试试规律，于是盲算一波，

2 时候 ： 1/(2/2)
4 的时候 ：2 * 3 / (4/2)
8 的时候 ：2 * 3 * 4 * 5 * 6 * 7 /(8/2)

n的时候 2 * 3 * 4 * 5 *… *(n-1)/(n/2)
然后 一试 20
确实是答案 交 对了；

#include <iostream>
#include<algorithm>
#include<map>
#include<set>
#define ll long long
using namespace std;
int main()
{
   
    int n;
    cin>>n;
    ll sum=1;
    for(int i=2;i<n;i++)
    {
   
        sum*=i;
    }
    n/=2;
    cout<<sum/n<<endl;
    return 0;
}