证明:http://www.cnblogs.com/linyujun/p/5194184.html#4010311
方法一:
复杂度O(logn)
inv(a) = a^(p-2) (mod p)
LL pow_mod(LL a, LL b, LL p){//a的b次方求余p
LL ret = 1;
while(b){
if(b & 1) ret = (ret * a) % p;
a = (a * a) % p;
b >>= 1;
}
return ret;
}
LL Fermat(LL a, LL p){//费马求a关于b的逆元
return pow_mod(a, p-2, p);
}
方法二:
#include
typedef long long LL;
void ex_gcd(LL a, LL b, LL &x, LL &y, LL &d){
if (!b) {d = a, x = 1, y = 0;}
else{
ex_gcd(b, a % b, y, x, d);
y -= x * (a / b);
}
}
LL inv(LL t, LL p){//如果不存在,返回-1
LL d, x, y;
ex_gcd(t, p, x, y, d);
return d == 1 ? (x % p + p) % p : -1;
}
int main(){
LL a, p;
while(~scanf("%lld%lld", &a, &p)){
printf("%lld\n", inv(a, p));
}
}
方法三:
#include
typedef long long LL;
LL inv(LL t, LL p) {//求t关于p的逆元,注意:t要小于p,最好传参前先把t%p一下
return t == 1 ? 1 : (p - p / t) * inv(p % t, p) % p;
}
int main(){
LL a, p;
while(~scanf("%lld%lld", &a, &p)){
printf("%lld\n", inv(a%p, p));
}
}