技術標籤：遞推DPYbtoj

D e s c r i p t i o n Description Description

I n p u t Input Input

一行，有兩個用空格隔開的整數n,m。

O u t p u t Output Output

1個整數，表示符合題意的方法數。

S a m p l e Sample Sample I n p u t Input Input

3 3

S a m p l e Sample Sample O u t p u t Output Output

2

H i n t Hint Hint

對於 40 % 40\% 40%的資料滿足， 3 ≤ n ≤ 30 , 1 ≤ m ≤ 20 3\leq n \leq 30, 1 \leq m \leq 20

T r a i n Train Train o f of of T h o u g h t Thought Thought

啊這
F i , j = F i − 1 , j − 1 + F i − 1 , j + 1 F_{i,j}=F_{i-1,j-1}+F_{i-1,j+1} Fi,j​=Fi−1,j−1​+Fi−1,j+1​

#include<algorithm>
#include
 
<iostream>
#include<cstring>
#include<cstdio>
#define ll long long 
using namespace std;

ll F[105][105];
ll n, m;

int main()
{
	scanf("%lld%lld", &n, &m);
	F[0][1] = 1;
	for(int i = 1; i <= m; ++i)
		for(int j = 1;j <= n; ++j)
		{
 			int x = (j - 1 >= 1) ?
 
 j - 1 : n;
			int y = (j + 1 <= n) ? j + 1 : 1;
			F[i][j] = F[i - 1][x] + F[i - 1][y];
		}
	printf("%lld", F[m][1]);
	return 0;
}