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FFT code This's my ancient code.:) |
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fixed the typo |
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If you want to calc $$$f(x) * g(x)$$$, you can simply calc $$$(f(x) + g(x)i)^2=f(x)^2 - g(x)^2 +2f(x)g(x)i$$$. |
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If you set the max(B) and min(B), the numbers between them can be chose or not, so the case number is $$$2 ^ {delta~index - 1}$$$(after sorting). Sorry, I I read it wrong.(C -> B Here is the explaination of the problem C: It's easy to see that we can separate different prime factors. And if $$$n = p^t$$$, we can iterate i from 1 to t(the number we choose), then $$$\dbinom t i \dbinom {n - 1}{i-1}$$$ is counted.(the first part we choose the set, the second part is the number of putting the $$$n$$$ numbers into $$$i$$$ indexes.) |
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+6
How about Atcoder DP Contest? |
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+6
It's scheduled at a nice time since I will go to school and I can't stay up late to compete at Codeforces. I'm a Chinese student. Thx for the great arragement. |
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