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Non-linear Thinking with CUDA

42 点作者 StylifyYourBlog超过 9 年前

2 条评论

rielfowler超过 9 年前
Why is the naive solution n^3? Generating one partial sum costs n, and you need to do it n times, so n^2. Even if you need to store all the intermediate sums, you get them as you sum the entire partial sum. No memoization needed. What am I missing here?
评论 #10793971 未加载
amelius超过 9 年前
I'm wondering if this problem can be approached as a standard convolution problem.
评论 #10794747 未加载