随笔分类 -  数学-数论-拓展中国定理

摘要:求: $$\Large\begin{cases}S\equiv b_1\pmod {a_1}\ S\equiv b_2\pmod {a_2}\ \cdots\ S\equiv b_i\pmod {a_i}\ \cdots\ S\equiv b_n\pmod {a_n}\ \end{cases}$$ 阅读全文
posted @ 2022-01-18 15:10 zhangtingxi 阅读(41) 评论(0) 推荐(0)