20240928模拟赛Agenius将模运算转化，\(\sum_{i=1}^{n}a_i\bmodk=\sum_{i=1}^{n}(a_i-\lfloor\frac{a_i}{k}\rfloor\timesk)=sum-k\sum_{i=1}^n\lfloor\frac{a_i}{k}\rfloor=s\)。移项得到\(sum-s=k\sum_{i=1}^n\lfloor\frac{a_i}{k}\rfloor\)。于是\(sum-s\)是