sunglassman

摘要： int phi[M], prime[M], all[M], cnt; void Euler() { phi[1] = 1; for (int i = 2; i < M; ++i) { if (!prime[i]) all[cnt++] = i, phi[i] = i - 1; for (int j 阅读全文

摘要： ll exgcd(ll a, ll b, ll c, ll& x, ll& y) { if (b == 0) { x = c, y = 0; //需要对ab作互素化 return a; } ll d = exgcd(b, a % b, c, y, x); y -= a / b * x; return 阅读全文