摘要: exgcd: pair<ll, ll> exgcd(ll a, ll b) { if (!b) return make_pair(1, 0); pair<ll, ll> x = exgcd(b, a % b); return make_pair(x.second, x.first - a / b * 阅读全文
posted @ 2020-12-27 21:53 Linshey 阅读(153) 评论(0) 推荐(0) 编辑