\[\sum_{i=1}^{n} \mu(i)\\ \sum_{i=1}^{n} \varphi(i)\\ \]
杜教筛模板：
\[S(n) = \sum_{i=1}^{n} f(i)\\ \sum_{i=1}^{n} \sum_{d|i} f(d) * g(\frac{i}{d})\\ =\sum_{i=1}^{n} g(i) \sum_{d=1}^{\lfloor \frac{n}{i}\rfloor} f(d)\\ =\sum_{i=1}^{n} g(i) S(\lfloor \frac{n}{i}\rfloor)\\ g(1)S(n) =\sum_{i=1}^{n} g(i) S(\lfloor \frac{n}{i}\rfloor) - \sum_{i=2}^{n} g(i) S(\lfloor \frac{n}{i}\rfloor) \]
具体分析：
\[S1(n) = \sum_{i=1}^{n}\mu(i)\\ g = 1\\ g(1)S(n) =\sum_{i=1}^{n} g(i) S(\lfloor \frac{n}{i}\rfloor) - \sum_{i=2}^{n} g(i) S(\lfloor \frac{n}{i}\rfloor)\\ S1(n) =\sum_{i=1}^{n} \sum_{d|i}\mu(d) - \sum_{i=2}^{n} S1(\lfloor \frac{n}{i}\rfloor)\\ S1(n) =\sum_{i=1}^{n} [i =1] - \sum_{i=2}^{n} S1(\lfloor \frac{n}{i}\rfloor)\\ S1(n) =1 - \sum_{i=2}^{n} S1(\lfloor \frac{n}{i}\rfloor)\\ \]
\[S2(n) = \sum_{i=1}^{n}\phi(i)\\ g = 1\\ g(1)S(n) =\sum_{i=1}^{n} g(i) S(\lfloor \frac{n}{i}\rfloor) - \sum_{i=2}^{n} g(i) S(\lfloor \frac{n}{i}\rfloor)\\ S2(n) =\sum_{i=1}^{n} \sum_{d|i}\phi(d) - \sum_{i=2}^{n} S2(\lfloor \frac{n}{i}\rfloor)\\ S2(n) =\sum_{i=1}^{n} i- \sum_{i=2}^{n} S2(\lfloor \frac{n}{i}\rfloor)\\ S2(n) =\frac{n*(n + 1)}{2} - \sum_{i=2}^{n} S2(\lfloor \frac{n}{i}\rfloor)\\ \]
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。