洛绫璃

摘要： #题意 $d_1, d_2$是 a 的因子, 找到 gcd(\(d_ 1 + d_2\), a) = 1 #题解 gcd(x, y) = gcd(x + y, y) = gcd(x, y + x) = gcd(x, \(y^n\)) = gcd(\(x^n\), y) = gcd(x + y, x 阅读全文