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比较慢(16ms)的公式: f(n, m)=f(n-2^k, m)+f(n-2^(k+1), m-2^(k+1))

Posted by lddlinan at 2020-09-17 13:43:46 on Problem 1430 
   f(n, m)
=  m*f(n-1, m)+f(n-1,m-1)
=  m*(m*f(n-2, m)+f(n-2, m-1))+(m-1)*f(n-2, m-1)+f(n-2, m-2)
=  m*f(n-2, m) + f(n-2, m-1)+ f(n-2, m-2)
=  f(n-1, m) + f(n-2, m-2)
=  f(n-2, m) + f(n-3, m-2) + f(n-3, m-2) + f(n-4, m-4)
=  f(n-2, m) + f(n-4, m-4)
=  f(n-4, m) + f(n-8, m-8)
...
=  f(n-2^k, m) + f(n-2^(k+1), m-2^(k+1))
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