2018-08-19发表2018-08-19更新acm模板几秒读完 (大约100个字)0次访问逆元模板求逆元模板递推求逆元12345int inv[maxn];int inv[1] = 1;rep(i,2,max){ inv[i] = inv[mod%i]*(mod-mod/i)%mod;} 费马小定理求逆元12345678910111213141516ll extend_gcd(ll a,ll b,ll x,ll y){ if(a==0 && b==0) return -1; if(b==0) { x = 1;y = 0; return a; } ll d = extend_gcd(b,a%b,y,x); y -= a/b*x; return d;}ll mod_reverse(ll a,ll n){ ll x,y; ll d = extend_gcd(a,n,x,y); if(d==1) return (x%n+n)%n; else return -1;} 逆元模板https://www.cheasim.com/acm%E6%A8%A1%E6%9D%BF/2018/08/19/%E9%80%86%E5%85%83%E6%A8%A1%E6%9D%BF.html作者CheaSim发布于2018-08-19更新于2018-08-19许可协议