摘要：
思路： 已知：A%9973 = n; gcd(B,9973)=1; 求（A/B)%9973; k = (A/B)%9973 - > A/B - 9973*y = k; -> A = k*B+9973*y; 带入 A%9973 = n中得 k*B%9973 = n; kB - 9973*y = n; 阅读全文
摘要：
参考博客： https://blog.csdn.net/u012469987/article/details/39041797 https://blog.csdn.net/qq_22902423/article/details/50569835 代码： #include<iostream> #inc 阅读全文