数论：求逆元的三种模板

1.线性求逆元

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typedef long long ll
int n;
int mode;
ll inv[maxn];

void init(int mode){//线性求逆元
    inv[0]=inv[1]=1;
    for (int i=2;i<maxn;i++){
        inv[i] = ((mode-mode/i)*inv[mode % i]) % mode;
    }
}

2.费马小定理，快速幂求逆元

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ll mypow(ll a,ll b){//快速幂
    ll ret=1;
    while(b){
        if(b&1){
            ret=ret*a%mode;
        }
        a=a*a%mode;
        b>>=1;
    }
    return ret;
}

ll C_inv(ll x){//快速幂——费马小定理求逆元
    return mypow(x,mode-2);
}

3.扩展欧几里得求逆元

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void gcd(ll a,ll b,ll &d,ll &x,ll &y) {
    if(!b) { d=a; x=1; y=0;}
    else {gcd(b,a%b,d,y,x); y -= x*(a/b);}
}

ll gcd_inv(ll a,ll n) {//扩展欧几里得求逆元
    ll d,x,y;
    gcd(a,n,d,x,y);
    return d == 1 ? (x+n) % n : -1;
}

4.java版高精求快速幂和逆元：

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import java.math.*;
import java.util.*;

public class Main{
	static BigInteger a,b;
	static BigInteger Mod = BigInteger.valueOf(19260817);
	
	static BigInteger mypow(BigInteger a,BigInteger b) {

		BigInteger ret = BigInteger.valueOf(1);
		while(0 != b.compareTo(BigInteger.valueOf(0))) {
			if(0 == b.mod(BigInteger.valueOf(2)).compareTo(BigInteger.valueOf(1))){
				ret = ret.multiply(a).mod(Mod);
			}
			a = a.multiply(a).mod(Mod);
			b=b.divide(BigInteger.valueOf(2));
		}
		return ret;
	}
	static BigInteger inv(BigInteger x) {
		BigInteger Inv = mypow(x,Mod.subtract(BigInteger.valueOf(2)));
		return Inv;
	}
	public static void main(String[] args) {
		Scanner cin = new Scanner (System.in);
		a=cin.nextBigInteger();
		b=cin.nextBigInteger();
		BigInteger ans = a.multiply(inv(b)).mod(Mod);
		System.out.println(ans);
	}
}