鏼爷的冬令营课件
核猩公式
[ln (1+x)=x-frac{1}{2}x^2+frac{1}{3}x^3-frac{1}{4}x^4+...
]
版本1
[egin{aligned}
prod_{i=1}^{n}(1+x^i+x^{2i}+...)^{a_i}=&prod_{i=1}^{n}(frac{1}{1-x^i})^{a_i}\
=&exp(sum_{i=1}^{n}-a_iln(1-x^i))\
=&exp(sum_{i=1}^{n}a_isum_{j=1}^{+infty}frac{x^{ij}}{j}) (其实感觉推到这步就能做了的说)\
=&exp(sum_{j=1}^{+infty}frac{1}{j}sum_{i=1}^{n}a_ix^{ij})\
=&exp(sum_{j=1}^{+infty}frac{1}{j}A(x^j))
end{aligned}
]
可以(A(x))有用的项一共只有(sum_{i=1}^{n}lfloorfrac{n}{i} floor=O(n log n))项。
版本2
[egin{aligned}
prod_{i=1}^{n}(1+x^i)^{a_i}=&exp(sum_{i=1}^{n}a_iln(1+x^i))\
=&exp(sum_{i=1}^{n}a_isum_{j=1}^{+infty}(-1)^{j+1}frac{x^{ij}}{j})\
=&exp(sum_{j=1}^{+infty}frac{(-1)^{j+1}}{j}sum_{i=1}^{n}a_ix^{ij})\
=&exp(sum_{j=1}^{+infty}frac{(-1)^{j+1}}{j}A(x^j))
end{aligned}
]
版本3
[egin{aligned}
prod_{i=1}^{n}(1+x^i+x^{2i}+...+x^{a_ii})=&prod_{i=1}^{n}frac{1-x^{a_ii}}{1-x^i}\
=&exp(sum_{i=1}^{n}(ln(1-x^{a_ii})-ln(1-x^i)))\
=&exp(sum_{i=1}^{n}(sum_{j=1}^{+infty}frac{x^{ij}}{j})-(sum_{j=1}^{+infty}frac{x^{a_iij}}{j}))
end{aligned}
]