rt，十分容易推出本题的转移方程为： f(i,j)=∑k=0a[i]f(i−1,j−k)f(i, j) = \sum_{k=0}^{a[i]}f(i - 1, j - k)f(i,j)=∑k=0a[i]f(i−1,j−k) 但是进一步想f(i,j−1)=∑k=1a[i]f(i−1,j−1−k)f(i, j - 1) = \sum_{k=1}^{a[i]}f(i - 1, j - 1 - k)f(i,j−1)=∑k=1a[i]f(i−1,j−1−k),其仅仅比f(i,j)=∑k=1a[i]f(i−1,j−k)f(i, j) = \sum_{k=1}^{a[i]}f(i - 1, j - k)f(i,j)=∑k=1a[i]f(i−1,j−k)多了f(i−1,j−1−a[i])f(i - 1,j - 1 - a[i])f(i−1,j−1−a[i]) 那么是否可以得到f(i,j)=f(i−1,j)+f(i,j−1)−f(i−1,j−1−a[i])f(i, j) = f(i - 1, j)+f(i , j-1)-f(i - 1 , j - 1 - a[i])f(i,j)=f(i−1,j)+f(i,j−1)−f(i−1,j−1−a[i])呢？ 如果此问题太若只请别喷，谢谢
