ユーザー解説 by dama_math
つねに ⌈i+1j⌉−⌊ij⌋=1\left\lceil \frac{i+1}{j} \right\rceil-\left\lfloor \frac{i}{j} \right\rfloor=1⌈ji+1⌉−⌊ji⌋=1 が成り立つから, (∑i=1106+1∑j=1106+1⌈ij⌉)−(∑i=1106∑j=1106⌊ij⌋)=∑i=1106∑j=1106(⌈i+1j⌉−⌊ij⌋)+∑j=1106+1⌈1j⌉+∑i=1106⌈i+1106+1⌉=(106)2+(106+1)+106=1000002000001.\begin{aligned} \Biggl(\sum_{i=1}^{10^6+1}\sum_{j=1}^{10^6+1}{\left\lceil \frac{i}{j} \right\rceil}\Biggr)-\Biggl(\sum_{i=1}^{10^6}\sum_{j=1}^{10^6}{\left\lfloor \frac{i}{j} \right\rfloor}\Biggr) &=\sum_{i=1}^{10^6}\sum_{j=1}^{10^6}\Biggl(\left\lceil\frac{i+1}{j} \right\rceil-\left\lfloor \frac{i}{j}\right\rfloor\Biggr)+\sum_{j=1}^{10^6+1}{\left\lceil \frac{1}{j} \right\rceil}+\sum_{i=1}^{10^6}\left\lceil \frac{i+1}{10^6+1} \right\rceil\\ &=(10^6)^2+(10^6+1)+10^6\\ &=\textbf{1000002000001}. \end{aligned}(i=1∑106+1j=1∑106+1⌈ji⌉)−(i=1∑106j=1∑106⌊ji⌋)=i=1∑106j=1∑106(⌈ji+1⌉−⌊ji⌋)+j=1∑106+1⌈j1⌉+i=1∑106⌈106+1i+1⌉=(106)2+(106+1)+106=1000002000001.