【20180318】2018北京集訓測試賽（六）
阿新 • • 發佈：2018-03-18
sum cnblogs 中間 com phi gpo 但是 分享圖片 problem
\[ =\sum_{i=1}^{n}\sum_{d|i}d\sum_{j=1}^{\left \lfloor \frac{i}{d} \right \rfloor}\sum_{k=1}^{\left \lfloor \frac{i}{d} \right \rfloor}\sum_{e|\frac{i}{d},e|j,e|k}\mu(e) \]
\[ =\sum_{i=1}^{n}\sum_{e=1}^{i}\mu(e)\sum_{d|i,d|\frac{i}{d}}d\sum_{e|j}\sum_{e|k} \]
\[ =\sum_{i=1}^{n}\sum_{e=1}^{i}\mu(e)\sum_{d|i,ed|i}d\left \lfloor \frac{i}{de} \right \rfloor^2 \]
\[ 設 t=de \]
\[ =\sum_{i=1}^{n}\sum_{t|i}\sum_{d|t}d\mu(\frac{t}{d})\left \lfloor \frac{i}{t} \right \rfloor^2 \]
\[ =\sum_{i=1}^{n}\sum_{t|i}\varphi(t)\left \lfloor \frac{i}{t} \right \rfloor^2 \]
交換一下（不交換也可以，但是結果的求和項會反過來
\[ =\sum_{i=1}^{n}\sum_{t|i}\varphi(\left \lfloor \frac{i}{t} \right \rfloor)t^2 \]
\[ =\sum_{t=1}^{n}t^2\sum_{t|i}\varphi(\left \lfloor \frac{i}{t} \right \rfloor) \]
\[ =\sum_{t=1}^{n}t^2\sum_{i=1}^{\left \lfloor \frac{n}{t} \right \rfloor}\varphi(t) \]
……就這樣，中間沒有交換結果反過來了，還以為發現了什麽驚天大秘密（像個智障
菜雞滾回石家莊了233
Problem B: 求和
題解&反思：
好久沒寫反演了真刺激
大力推公式就好咯
\[
\sum_{i=1}^{n}\sum_{j=1}^{i}\sum_{k=1}^{i}gcd(i,j,k)
\]
\[
=\sum_{i=1}^{n}\sum_{d|i}d\sum_{j=1}^{i}\sum_{k=1}^{i}[gcd(i,j,k)==d]
\]
\[
=\sum_{i=1}^{n}\sum_{d|i}d\sum_{j=1}^{\left \lfloor \frac{i}{d} \right \rfloor}\sum_{k=1}^{\left \lfloor \frac{i}{d} \right \rfloor}[gcd(\frac{i}{d},j,k)==1]
\]
\[ =\sum_{i=1}^{n}\sum_{d|i}d\sum_{j=1}^{\left \lfloor \frac{i}{d} \right \rfloor}\sum_{k=1}^{\left \lfloor \frac{i}{d} \right \rfloor}\sum_{e|\frac{i}{d},e|j,e|k}\mu(e) \]
\[ =\sum_{i=1}^{n}\sum_{e=1}^{i}\mu(e)\sum_{d|i,d|\frac{i}{d}}d\sum_{e|j}\sum_{e|k} \]
\[ =\sum_{i=1}^{n}\sum_{e=1}^{i}\mu(e)\sum_{d|i,ed|i}d\left \lfloor \frac{i}{de} \right \rfloor^2 \]
\[ 設 t=de \]
\[ =\sum_{i=1}^{n}\sum_{t|i}\sum_{d|t}d\mu(\frac{t}{d})\left \lfloor \frac{i}{t} \right \rfloor^2 \]
\[ =\sum_{i=1}^{n}\sum_{t|i}\varphi(t)\left \lfloor \frac{i}{t} \right \rfloor^2 \]
交換一下（不交換也可以，但是結果的求和項會反過來
\[ =\sum_{i=1}^{n}\sum_{t|i}\varphi(\left \lfloor \frac{i}{t} \right \rfloor)t^2 \]
\[ =\sum_{t=1}^{n}t^2\sum_{t|i}\varphi(\left \lfloor \frac{i}{t} \right \rfloor) \]
\[ =\sum_{t=1}^{n}t^2\sum_{i=1}^{\left \lfloor \frac{n}{t} \right \rfloor}\varphi(t) \]
……就這樣，
【20180318】2018北京集訓測試賽（六）