Codeforces Gym 100015B Ball Painting 找规律

Ball Painting

题目连接：

http://codeforces.com/gym/100015/attachments

Description

There are 2N white balls on a table in two rows, making a nice 2-by-N rectangle. Jon has a big paint bucket
full of black paint. (Don’t ask why.) He wants to paint all the balls black, but he would like to have some
math fun while doing it. (Again, don’t ask why.) First, he computed the number of di!erent ways to paint
all the balls black. In no time, he ﬁgured out that the answer is (2N)! and thought it was too easy. So, he
introduced some rules to make the problem more interesting.

• The ﬁrst ball that Jon paints can be any one of the 2N balls.
• After that, each subsequent ball he paints must be adjacent to some black ball (that was already

painted). Two balls are assumed to be adjacent if they are next to each other horizontally, vertically,
or diagonally.

Jon was quite satisﬁed with the rules, so he started counting the number of ways to paint all the balls
according to them. Can you write a program to ﬁnd the answer faster than Jon?

Input

The input consists of multiple test cases. Each test case consists of a single line containing an integer N,
where 1 ! N ! 1,000. The input terminates with a line with N = 0. For example:
1
2
3
0

Output

For each test case, print out a single line that contains the number of possible ways that Jon can paint all
the 2N balls according to his rules. The number can become very big, so print out the number modulo
1,000,000,007. For example, the correct output for the sample input above would be:
2
24
480

Sample Input

Sample Output

480

Hint

题意

给你2*n的矩阵，都是白色

1.一开始你随便选一个染成黑色

2.接下来你可以选择一个和黑色相邻的白色，染成黑色

问你有多少种染法

题解：

我是OEIS的，蛤蛤

正常做可以DP

代码

#include<bits/stdc++.h>
using namespace std;
const int mod = 1e9+7;
//(4-8*n)*a(n)+a(n+1) =0.
long long a[1200];
void pre()
{
    a[1]=1;
    a[2]=2;
    for(int i=3;i<=1100;i++)
        a[i]=(a[i-1]*(8LL*(i-1LL)-4LL))%mod;
}
long long n;
int main()
{
    pre();
    while(cin>>n)
    {
        if(n==0)break;
        cout<<a[n+1]<<endl;
    }
}