摘要: 容易发现答案为 ∑i=1b[gcd⁡(i,n)=1]−∑i=1a−1[gcd⁡(i,n)=1]\sum \limits_{i=1}^b [\gcd(i,n)=1]-\sum \limits_{i=1}^{a-1} [\gcd(i,n)=1]i=1∑b​[gcd(i,n)=1]−i=1∑a−1​[gc 阅读全文
posted @ 2023-07-08 11:20 HappyBobb 阅读(20) 评论(0) 推荐(0)