不难发现 j↦f(i,j)j\mapsto f(i,j)j↦f(i,j) 取遍自然数且单调不降。记 q(i,j)=max{k∣f(i,k)=j}q(i,j)=\max\{k|f(i,k)=j\}q(i,j)=max{k∣f(i,k)=j}，则由定义 q(i,j)=1+q(i−1,j−1)+q(i,j−1)q(i,j)=1+q(i-1,j-1)+q(i,j-1)q(i,j)=1+q(i−1,j−1)+q(i,j−1)
归纳可得 j≥ij\ge ij≥i 时 q(i,j)=2i−1q(i,j)=2^i-1q(i,j)=2i−1，所以 q(7,100)=127,q(6,100)=63q(7,100)=127,q(6,100)=63q(7,100)=127,q(6,100)=63，所以 f(100,100)=7f(100,100)=7f(100,100)=7
取 min 的次数 w(i,j)=2w(i,j−1)+w(i−1,j−1)+1w(i,j)=2w(i,j-1)+w(i-1,j-1)+1w(i,j)=2w(i,j−1)+w(i−1,j−1)+1
没问题吧 QwQ
