hihoCoder 1596 : Beautiful Sequence

Description

Consider a positive integer sequence a[1], ..., a[n] (n ≥ 3). If for every 2 ≤ i ≤ n-1, a[i-1] + a[i+1] ≥ 2 × a[i] holds, then we say this sequence is beautiful.

Now you have a positive integer sequence b[1], ..., b[n]. Please calculate the probability P that the resulting sequence is beautiful after uniformly random shuffling sequence b.

You're only required to output (P × (n!)) mod 1000000007. (Obviously P × (n!) is an integer)

Input

First line contains an integer n. (3 ≤ n ≤ 60)

Following n lines contain integers b[1], b[2], ..., b[n]. (1 ≤ b[i] ≤ 1000000000)

Output

Output (P × (n!)) mod 1000000007.

Sample Input

4
1
2
1
3

Sample Output

8
https://hihocoder.com/problemset/problem/1596
dp鬼题
题目条件可化为a[i+1]-a[i]>=a[i]-a[i-1]
考虑排序再做分配
根据分析我们发现最后的高度序列是一个勾函数，先减小后增大
我们讨论b[i-1],b[i],b[i+1]的情况
显然i-1和i+1不可能同时大于i
只可能一个大于i一个小于，或两个都大于
但是两个都大于的情况显然只有一次，两边是不会有的
如5 3 5 1 3 5
那么就出现了两个都小于的情况
那么我们就可以dp
令f[i][j][k][l]表示最左边两个为i,j 最右边两个为k,l
我们先将最小值放入f[1][0][1][0]=1
接下来要放的数为max(i,k)+1,为什么？因为是排过序的，从小到大放就行了
判断是否满足：a[i+1]-a[i]>=a[i]-a[i-1]
还有一个细节：
当有多个最小值时，显然不能直接dp，以最小值数量为l=3举例
因为直接dp只能得到4种，而实际有6种(此题鬼处)
所以把所有最小值缩为一个，最后乘l!

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int Mod=1000000007;
long long f[61][61][61][61],a[61],p[61],tmp,ans;
int n;
int main()
{int i,j,k,l=0;
    cin>>n;
    for (i=1;i<=n;i++)
    {
        scanf("%lld",&a[i]);
    }
    sort(a+1,a+n+1);
    p[0]=1;
    for (i=1;i<=n;i++)
    p[i]=(p[i-1]*i)%Mod;
    for (i=1;i<=n;i++)
    if (a[i]==a[1]) l++;
    tmp=p[l];
    for (i=2;i<=n-l+1;i++)
    a[i]=a[i+l-1];
    n=n-l+1;
    f[1][0][1][0]=1;
    for (i=1;i<=n;i++)
    {
        for (j=0;j<=n-1;j++)
        {
            for (k=1;k<=n;k++)
            {
                for (l=0;l<=n-1;l++)
                {
                    int z=max(i,k)+1;
                    if (z==n+1)
                    {
                        ans+=f[i][j][k][l];
                        ans%=Mod;
                        continue;
                    }
                    if (a[z]-a[k]>=a[k]-a[l]||l==0)
                    f[i][j][z][k]+=f[i][j][k][l],f[i][j][z][k]%=Mod;
                    if (a[z]-a[i]>=a[i]-a[j]||j==0)
                    f[z][i][k][l]+=f[i][j][k][l],f[z][i][k][l]%=Mod;
                }
            }
        }
    }
    cout<<(ans*tmp)%Mod;
}