[sum_{i=1}^{n} mu(i)\
sum_{i=1}^{n} varphi(i)\
]
杜教筛模板：
[S(n) = sum_{i=1}^{n} f(i)\
sum_{i=1}^{n} sum_{d|i} f(d) * g(frac{i}{d})\
=sum_{i=1}^{n} g(i) sum_{d=1}^{lfloor frac{n}{i}
floor} f(d)\
=sum_{i=1}^{n} g(i) S(lfloor frac{n}{i}
floor)\
g(1)S(n) =sum_{i=1}^{n} g(i) S(lfloor frac{n}{i}
floor) - sum_{i=2}^{n} g(i) S(lfloor frac{n}{i}
floor)
]
具体分析：
[S1(n) = sum_{i=1}^{n}mu(i)\
g = 1\
g(1)S(n) =sum_{i=1}^{n} g(i) S(lfloor frac{n}{i}
floor) - sum_{i=2}^{n} g(i) S(lfloor frac{n}{i}
floor)\
S1(n) =sum_{i=1}^{n} sum_{d|i}mu(d) - sum_{i=2}^{n} S1(lfloor frac{n}{i}
floor)\
S1(n) =sum_{i=1}^{n} [i =1] - sum_{i=2}^{n} S1(lfloor frac{n}{i}
floor)\
S1(n) =1 - sum_{i=2}^{n} S1(lfloor frac{n}{i}
floor)\
]
[S2(n) = sum_{i=1}^{n}phi(i)\
g = 1\
g(1)S(n) =sum_{i=1}^{n} g(i) S(lfloor frac{n}{i}
floor) - sum_{i=2}^{n} g(i) S(lfloor frac{n}{i}
floor)\
S2(n) =sum_{i=1}^{n} sum_{d|i}phi(d) - sum_{i=2}^{n} S2(lfloor frac{n}{i}
floor)\
S2(n) =sum_{i=1}^{n} i- sum_{i=2}^{n} S2(lfloor frac{n}{i}
floor)\
S2(n) =frac{n*(n + 1)}{2} - sum_{i=2}^{n} S2(lfloor frac{n}{i}
floor)\
]
其中:
[sum_{d|n}^{n} mu(d) = [n = 1]\
sum_{d|n}^{n} phi(d) = n\
]
上面的式子前半部分可以直接得出，后面的用数论分块+递归解决。