证明
∀gcd(a,b)=1,∑i=1a−1⌊bia⌋=∑i=1b−1⌊aib⌋\forall \gcd(a, b)=1, \sum_{i=1}^{a-1}\left \lfloor \frac{bi}{a} \right \rfloor=\sum_{i=1}^{b-1}\left \lfloor \frac{ai}{b} \right \rfloor∀gcd(a,b)=1,∑i=1a−1⌊abi⌋=∑i=1b−1⌊bai⌋
