摘要:
$ ax + by = c $方程的通解: 令 $ g = gcd(a, b) $ 设有一组特解 $x_0, y_0$ 是 $a' x + b' y = 1$的特解 $ x = x_0 \frac{c}{g} + k \frac{b}{g} $ $y = y_0 \frac{c}{g} - k \f 阅读全文
摘要:
#include<bits/stdc++.h> #warning Tab and SPace !!! using namespace std; const int N = 500+10, V = 50 + 5, L = 1000 + 10; enum{C_root, C_prevar, C_set, 阅读全文