109th LeetCode Weekly Contest Knight Dialer

A chess knight can move as indicated in the chess diagram below:

 .           

This time, we place our chess knight on any numbered key of a phone pad (indicated above), and the knight makes N-1 hops.  Each hop must be from one key to another numbered key.

Each time it lands on a key (including the initial placement of the knight), it presses the number of that key, pressing N digits total.

How many distinct numbers can you dial in this manner?

Since the answer may be large, output the answer modulo 10^9 + 7.

Example 1:

Input: 1
Output: 10

Example 2:

Input: 2
Output: 20

Example 3:

Input: 3
Output: 46

Note:

• 1 <= N <= 5000

说就是一个马在键盘上跳，能得到多少种的数字，比如1就是不跳。。在原地就是0到9,10个数

那么可以看看，0这个位置只有4和6可以跳过来，1是6,8可以跳过来，5是不可能到的，所以就有下面的代码（参考别人的，自己写的不好看）

class Solution {
public:
    int mod = ;
    int dp[],dp2[];
    int knightDialer(int N) {
       for(int i=;i<=;i++){
            dp[i]=;
        }
        for(int i=;i<N;i++){
            dp2[] = (dp[]+dp[])%mod;
            dp2[] = (dp[]+dp[])%mod;
            dp2[] = (dp[]+dp[])%mod;
            dp2[] = (dp[]+dp[])%mod;
            dp2[] = ((dp[]+dp[])%mod+dp[])%mod;
            dp2[] = ;
            dp2[] = ((dp[]+dp[])%mod+dp[])%mod;
            dp2[] = (dp[]+dp[])%mod;
            dp2[] = (dp[]+dp[])%mod;
            dp2[] = (dp[]+dp[])%mod;
            for(int j=;j<=;j++){
                dp[j] = dp2[j];
            }
        }
        int result = ;
        for(int i=;i<=;i++){
            cout<<dp[i]<< " ";
            result += dp[i] %mod;
            result %= mod;
        }
        cout<<endl;
        return result;
    }
};