T1
分别对序列和值域分块。
只需要做到(O(sqrt{n}))查询(O(1))修改就可以了。
这样的话与处理一下序列和值域分块的情况。
查询的时候动态的处理散块就可以在(O(sqrt{n}))复杂度维护(K)大值。
T2
直接推式子。
[egin{aligned}
ans&=sumlimits_{i=1}^{n}sumlimits_{j=1}^{n}sumlimits_{k=1}^{K}f_k(gcd(i,j))\
&=sumlimits_{k=1}^{K}sumlimits_{i=1}^{n}sumlimits_{j=1}^{n}f_k(gcd(i,j))\
&=sumlimits_{k=1}^{K}sumlimits_{d=1}^{n}f_k(d)sumlimits_{i=1}^{frac{n}{d}}sumlimits_{j=1}^{frac{n}{d}}[gcd(i,j)=1]\
&=sumlimits_{k=1}^{K}sumlimits_{d=1}^{n}f_k(d)sumlimits_{i=1}^{frac{n}{d}}sumlimits_{j=1}^{frac{n}{d}}sumlimits_{g|gcd(i,j)}mu(g)\
&=sumlimits_{k=1}^{K}sumlimits_{d=1}^{n}f_k(d)sumlimits_{g=1}^{frac{n}{d}}mu(g)lfloorfrac{n}{dg}
floor^2\
&=sumlimits_{k=1}^{K}sumlimits_{T=1}^{n}lfloorfrac{n}{T}
floor^2sumlimits_{d|T}mu(d)f_k(frac{T}{d})\
H_k(n)&=sumlimits_{d|n}mu(d)f_k(frac{n}{d})\
H_k&=mu*f_k\
H_k(p^e)&=mu(p^0)f_k(p^e)+mu(p^1)f_k(p^{e-1})\
n=&prodlimits_{i=1}^{w}p_i^{c_i}\
g(n)&=prodlimits_{i=1}^{n}(-1)^{c_i}\
end{aligned}]
[egin{aligned}
ans&=sumlimits_{i=1}^{n}sumlimits_{j=1}^{n}sumlimits_{k=1}^{K}f_k(gcd(i,j))\
&=sumlimits_{k=1}^{K}sumlimits_{i=1}^{n}sumlimits_{j=1}^{n}f_k(gcd(i,j))\
&=sumlimits_{k=1}^{K}sumlimits_{d=1}^{n}f_k(d)sumlimits_{i=1}^{frac{n}{d}}sumlimits_{j=1}^{frac{n}{d}}[gcd(i,j)=1]\
&=sumlimits_{k=1}^{K}sumlimits_{d=1}^{n}f_k(d)left(left(sumlimits_{i=1}^{frac{n}{d}}2varphi(i)
ight)-1
ight)\
res&=sumlimits_{k=1}^{K}sumlimits_{d=1}^{n}f_k(d)\
F_k(n)&=sumlimits_{i=1}^{n}f_k(i)\
&=sumlimits_{i=1}^{n}g(i)-sumlimits_{i=2}^{n}(-mu(i))sumlimits_{j=1}^{frac{n}{i^{k+1}}}g(i^{k+1}j)\
&=sumlimits_{i=1}^{n}mu(i)sumlimits_{j=1}^{frac{n}{i^{k+1}}}g(i^{k+1}j)\
&=sumlimits_{i=1}^{sqrt[k+1]{n}}mu(i)g^{k+1}(i)sumlimits_{j=1}^{frac{n}{i^{k+1}}}g(j)\
G(n)&=sumlimits_{i=1}^{n}g(i)\
g*delta&=eta\
sumlimits_{d|n}g(frac{n}{d})&=[exist iin N^{*},i^2=n]\
g*I&=[exist iin N^{*},i^2=n]\
sumlimits_{i=1}^{n}[exist jin N^{*},j^2=i]&=sumlimits_{i=1}^{n}sumlimits_{d|i}g(frac{i}{d})\
&=sumlimits_{d=1}^{n}sumlimits_{i=1}^{frac{n}{d}}g(i)\
&=G(n)+sumlimits_{d=2}^{n}G(frac{n}{d})\
G(n)&=sqrt{n}-sumlimits_{d=2}^{n}G(frac{n}{d})\
end{aligned}
]
杜教筛就完事了。
T3
对于两条序列，有如下结论和证明。
[a
ightarrow b,c
ightarrow d
]
[a
ightarrow d,c
ightarrow b
]
[(a,b)cap(c,d)=e,(a,e)+(e,b)+(c,e)+(e,d)=(a,d)+(c,b)
]
[(a,b)cap(c,d)=emptyset,rt,LCA(a,b),LCA(c,d)
]
这样我们只需要找到每个点当了多少次起点或者终点，然后按照最小字典序组合起来就可以了。
[fr=LCA(a,b),se=LCA(c,d)
]
[(a,fr)+(fr,b)+(c,se)+(se,d)=(a,fr)+(fr,se)+(se,d)+(c,se)+(se,fr)+(fr,b)
]