本文的部分latex垮掉了。我也不知道为什么
（有毒），建议将代码拷贝至本地md编编辑器阅读。
数论（整除）分块
整除分块：设函数(f(x)=lfloorfrac{N}{x} floor, xin N^+)，期望在根号时间内求得(sum_{x} f(x))的值。
【定理1】(f(x))的值域大小不超过(2sqrt{N})。
【定理2】若(x=q)为(f(x)=p)的解，则(x=lfloorfrac{N}{lfloor N/q floor} floor)是(f(x)=p)的最大解。
证明如下
[ ext{make } N=pq+r(0le r<q) \
ext{make } N=p*(q+k)+t(0le t<q+d) \
pk+t=r\
t=r-pkge 0 \
klelfloor frac{r}{p}
floor=lfloor frac{Nmod q}{p}
floor \
egin{aligned}
x_{max}&=q+k_{max} \
&=q+lfloor frac{Nmod q}{p}
floor\
&=q+lfloor frac{N-lfloor N/q
floor*q}{lfloor N/q
floor}
floor\
&=lfloor q+frac{N-lfloor N/q
floor*q}{lfloor N/q
floor}
floor\
&=lfloorfrac{lfloor N/q
floor*q +N-lfloor N/q
floor*q}{lfloor N/q
floor}
floor\
&=lfloorfrac{N}{lfloor N/q
floor}
floor\
end{aligned}
]
莫比乌斯函数
【定义】
定义 (mu(d)) 为
[mu(d)=egin{cases}
1&d=1\
(-1)^k&d=Pi^{k}_{i=1}P_i\
0&else
end{cases}
]
称 (mu(d)) 为关于 (d) 的莫比乌斯函数。
【性质】
- 对于任意 (nin N^*) ，有 (sum_{d|n}mu(d)=[n=1]) 。
- 对于任意 (nin N^*) ，有 (sum_{d|n}dfrac{mu(d)}{d}=dfrac{phi(n)}{n}) 。
【 递推】 结合线性筛素数。
【配套练习 】
[egin{aligned}
r&=sum_{pinPr}sum _{x=1}^nsum_{y=1}^m [gcd(x,y)=p]\
&=sum_{p=Pr}sum_{x=1}^{lfloor n/p
floor}sum_{y=1}^{lfloor m/p
floor}[gcd(x,y)=1] &star\
r&=sum_{p=Pr}sum_{x=1}^{lfloor n/p
floor}sum_{y=1}^{lfloor m/p
floor}sum_{d|gcd(x,y)}mu(d) &star\
&=sum_{p=Pr}sum_{d=1}^{lfloorfrac{min(n,m)}{p}
floor}mu(d)sum_{x=1}^{lfloor n/p
floor}sum_{y=1}^{lfloor m/p
floor}[d|gcd(x,y)]\
&=sum_{p=Pr}sum_{d=1}^{lfloorfrac{min(n,m)}{p}
floor}mu(d)lfloorfrac{n}{pd}
floorlfloorfrac{m}{pd}
floor\
ext{设 }k&=pd\
r&=sum_{p=Pr}sum_{d=1}^{lfloorfrac{min(n,m)}{p}
floor}mu(frac{k}{p})lfloorfrac{n}{k}
floorlfloorfrac{m}{k}
floor\
&=sum_{k=1}^{min(n,m)}sum_{pinPr,pmid k}mu(frac{k}{p})lfloorfrac{n}{k}
floorlfloorfrac{m}{k}
floor\
ext{预处理}f(k)&=sum_{pinPr,pmid k}mu(frac{k}{p}) ext{对于所有的 k(s)}\
r&=sum_{k=1}^{min(n,m)}f(k)lfloorfrac{n}{k}
floorlfloorfrac{m}{k}
floor\
end{aligned}
]
[egin{aligned}
r(n,m)&=sum_{i=1}^nsum_{j=1}^mfrac{i*j}{gcd(i,j)}\
&=sum_{d}sum_{i=1}^nsum_{j=1}^mfrac{i*j}{d} [gcd(i,j)=d]\
&=sum_{d}sum_{i=1}^{lfloorfrac{n}{d}
floor}sum_{j=1}^{lfloorfrac{m}{d}
floor}d*i*j[gcd(i,j)=1]\
&=sum_{d=1}^{min(n,m)}dsum_{i=1}^{lfloorfrac{n}{d}
floor}sum_{j=1}^{lfloorfrac{m}{d}
floor}i*j[gcd(i,j)=1]\
q(n,m)&=sum_{i=1}^nsum_{j=1}^m i*j[gcd(i,j)=1] &star\
&=sum_{i=1}^nsum_{j=1}^m i*jsum_{k|gcd(i,j)}mu(k) &star\
&=sum_{k=1}^{min(n,m)}mu(k)sum_{kmid i}^nsum_{kmid j}^m i*j\
&=sum_{k=1}^{min(n,m)}mu(k)sum_{i=1}^{lfloorfrac{n}{k}
floor}sum_{j=1}^{lfloorfrac{m}{k}
floor} i*j*k^2\
&=sum_{k=1}^{min(n,m)}k^2mu(k)sum_{i=1}^{lfloorfrac{n}{k}
floor}sum_{j=1}^{lfloorfrac{m}{k}
floor} i*j\
r(n,m)&=sum_{d=1}^{min(n,m)}d*q(d,lfloorfrac{n}{d}
floor,lfloorfrac{m}{d}
floor)
end{aligned}
]
莫比乌斯反演
【定理1】若 (F(n)) 和 (f(n)) 是定义在 (N) 上的两个函数，且满足 (F(n)=sum_{dmid n}f(d))，那么 (f(n)=sum_{dmid n}mu(d)F(dfrac{n}{d}))
证明：
[egin{aligned}
sum_{dmid n}mu(d)F(dfrac{n}{d})&=sum_{dmid n}mu(d)sum_{imiddfrac{n}{d}}f(i)\
&=sum_{imid n}f(i)sum_{dmiddfrac{n}{i}}mu(d)\
&=f(n) &star\
end{aligned}
]
【定理2】若 (F(n)) 和 (f(n)) 是定义在 (N) 上的两个函数，且满足 (F(n)=sum_{nmid d}f(d))，那么 (f(n)=sum_{nmid d}mu(dfrac{d}{n})F(d))
【配套练习】
[ ext{在忽略a的限制下}
egin{aligned}
F(x)&=sum_{dmid x} d \
r(n,m)&=sum_{i=1}^nsum_{j=1}^m F(gcd(i,j))\
&=sum_{d=1}^{min(n,m)}F(d)sum_{i=1}^nsum_{j=1}^m[gcd(i,j)=d]\
&=sum_{d=1}^{min(n,m)}F(d)sum_{d|k}mu(frac{k}{d})lfloorfrac{n}{k}
floorlfloorfrac{m}{k}
floor\
&=sum_{k=1}^{min(n,m)}lfloorfrac{n}{k}
floorlfloorfrac{m}{k}
floorsum_{dmid k}F(d)mu(frac{k}{d})
end{aligned}
]