Codeforces Round #175 (Div. 2)

D题， 纠结， 一开始想用组合数学的方法可是想不出怎么排列组合， 然后又准备暴力结果发现普通暴力在n=15的时候就不知道要运行多少时间。。。

于是去学习了一个 中途相遇的 搜索算法，快了很多， n=15的那组数据也可以在20s内跑完.

其实这个算法就是将 n^n 的复杂度变成了 2*n^(n/2) ,  可以快很多....

本质就是将整个搜索过程分成两半来搜。

 

D. Permutation Sum

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Permutation p is an ordered set of integers p₁,  p₂,  ...,  p_n, consisting of n distinct positive integers, each of them doesn't exceed n. We'll denote the i-th element of permutation p as p_i. We'll call number n the size or the length of permutation p₁,  p₂,  ...,  p_n.

Petya decided to introduce the sum operation on the set of permutations of length n. Let's assume that we are given two permutations of length n: a₁, a₂, ..., a_n and b₁, b₂, ..., b_n. Petya calls the sum of permutations a and b such permutation c of length n, where c_i = ((a_i - 1 + b_i - 1) mod n) + 1 (1 ≤ i ≤ n).

Operation  means taking the remainder after dividing number x by number y.

Obviously, not for all permutations a and b exists permutation c that is sum of a and b. That's why Petya got sad and asked you to do the following: given n, count the number of such pairs of permutations a and b of length n, that exists permutation c that is sum of a and b. The pair of permutations x, y (x ≠ y) and the pair of permutations y, x are considered distinct pairs.

As the answer can be rather large, print the remainder after dividing it by 1000000007 (10⁹ + 7).

Input

The single line contains integer n (1 ≤ n ≤ 16).

Output

In the single line print a single non-negative integer — the number of such pairs of permutations a and b, that exists permutation c that is sum of a and b, modulo 1000000007 (10⁹ + 7).

Sample test(s)

input

3

output

18

input

5

output

1800

#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;
#define MOD 1000000007
typedef __int64 LL;

int n;
int n1,n2;
int g1[20],g2[20];
int key;
int cnt1,cnt2;
LL sum1[1<<17],sum2[1<<17];

int pan(int i,int k)
{
    int sum=0;
    while(i)
    {
        sum += i&1;
        i=i>>1;;
    }
    return sum==k;
}

void dfs1(int s)
{
    if(s>n1)
    {
        sum1[key]++;
        return ;
    }
    for(int i=0;i<cnt1;i++)
    {
        int tmp=g1[i];
        if(g1[i]==-1) continue;
        int tmp1=(tmp+s-2)%n+1;
        if(key & ( 1<<(tmp1-1) ) ) continue;
        key |= ( 1<<(tmp1-1) );
        g1[i]=-1;
        dfs1(s+1);
        g1[i]=tmp;
        key ^= ( 1<<(tmp1-1) );
    }
}
void dfs2(int s)
{
    if(s>n)
    {
        sum2[key]++;
        return ;
    }
    for(int i=0;i<cnt2;i++)
    {
        int tmp=g2[i];
        if(g2[i]==-1) continue;
        int tmp1=(tmp+s-2)%n+1;
        if(key & ( 1<<(tmp1-1) ) ) continue;
        key |= ( 1<<(tmp1-1) );
        g2[i]=-1;
        dfs2(s+1);
        g2[i]=tmp;
        key ^= ( 1<<(tmp1-1) );
    }
}

int main()
{
    for(int i=1;i<=16;i++)
    {
        n=i;
        n1=n/2; n2=n-n1;
        LL ans=0;
        for(int i=0;i<=(1<<n)-1;i++)
        {
            if(pan(i,n1)==0) continue;
            cnt1=0; cnt2=0;
            for(int j=0;j<n;j++)
            {
                if( (1<<j)&i ) g1[cnt1++]=j+1;
                else g2[cnt2++]=j+1;
            }
            memset(sum1,0,sizeof(sum1));
            memset(sum2,0,sizeof(sum2));
            key=0; // 主要用来记录c的情况...
            dfs1(1);
            key=0;
            dfs2(n1+1);
            for(int j=0;j<=(1<<n)-1;j++)
            {
                int tmp=j^((1<<n)-1);
                ans=(ans+(sum1[j]*sum2[tmp])%MOD)%MOD;
            }
        }
        for(int i=1;i<=n;i++)
            ans=(ans*i)%MOD;
        printf("%I64d,",ans);
    }
    return 0;
}

/*#include <stdio.h>

int g[20]={0,1,0,18,0,1800,0,670320,0,734832000,0,890786230,0,695720788,0,150347555,0};

int main()
{
    int n;
    scanf("%d",&n);
    printf("%d",g[n]);
}*/