莫比乌斯反演学习笔记
余数求和 对于任意整数 \(x\in[1,n]\) 设 \(g(x)=\lfloor\frac{k}{\lfloor\frac{k}{x}\rfloor}\rfloor\) \(\because f(x)=\frac{k}{x}\) 单调递减 又 \((x)=\lfloor\frac{k}{\lfloor\frac{k}{x}\rfloor}\rfloor\geq\lfloor\frac{k}{(\frac{k}{x})}\rfloor=x\) \(\therefore \lfloor\frac{k}{g(x)}\rfloor\leq\lfloor\frac{k}{x}\rfloor\) \(\because \lfloor\frac{k}{g(x)}\rfloor\geq\lfloor\frac{k}{(\frac{k}{\lfloor\frac{k}{x}\rfloor})}\rfloor=\lfloor\frac{k}{k}\lfloor\frac{k}{x}\rfloor\rfloor=\lfloor\frac{k}{x}\rfloor\) \(\therefore \lfloor\frac{k}{g(x)}\rfloor=\lfloor\frac{k}{x}\rfloor\) 综上得,对于 \(i\in[x,\lfloor\frac{k}{\lfloor\frac{k}