摘要: 反演好难啊 qwq 。 ∑ d = 1 n d ( ∑ i = 1 n / d ∑ j = 1 m / d i ∗ j ∗ [ gcd ⁡ ( i , j ) = = 1 ] ) \sum_{d=1}^nd(\sum_{i=1}^{n/d}\sum_{j=1}^{m/d}i*j*[\gcd(i,j) 阅读全文
posted @ 2021-08-31 22:12 仰望星空的蚂蚁 阅读(11) 评论(0) 推荐(0)