【BZOJ-4522】密钥破解 数论 + 模拟 （ Pollard_Rho分解 + Exgcd求逆元 + 快速幂 + 快速乘）

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cstdlib>
using namespace std;
long long read()
{
    long long x=0,f=1; char ch=getchar();
    while (ch<‘0‘ || ch>‘9‘) {if (ch==‘-‘) f=-1; ch=getchar();}
    while (ch>=‘0‘ && ch<=‘9‘) {x=x*10+ch-‘0‘; ch=getchar();}
    return x*f;
}
long long e,N,c,r,P,Q;
long long Quick_Mul(long long x,long long y,long long p)
{
    long long re=0;
    for (long long i=y; i; i>>=1,x=(x+x)%p)
        if (i&1) re=(re+x)%p;
    return re;
}
long long Quick_Pow(long long x,long long y,long long p)
{
    long long re=1;
    for (long long i=y; i; i>>=1,x=Quick_Mul(x,x,p))
        if (i&1) re=Quick_Mul(re,x,p);
    return re;
}
void Exgcd(long long a,long long b,long long &x,long long &y)
{
    if (b==0) {x=1; y=0; return;}
    Exgcd(b,a%b,y,x); y-=(a/b)*x;
}
long long GetInv(long long n,long long p)
{
    long long x,y;
    Exgcd(n,p,x,y);
    return (x%p+p)%p;
}
long long Gcd(long long a,long long b)
{
    if (b==0) return a; 
    return Gcd(b,a%b);
}
#define T 10007
long long Pollard_Rho(long long n)
{
    long long x,y,cnt=1,k=2;
    x=rand()%(n-1)+1; y=x;
    while (1)
        {
            cnt++;
            x=(Quick_Mul(x,x,n)+T)%n;
            long long gcd=Gcd(abs(x-y),n);
            if (1<gcd && gcd<n) return gcd;
            if (x==y) return n;
            if (cnt==k) y=x,k<<=1;
        }
}
int main()
{
    srand(T);
    e=read(),N=read(),c=read();
    P=Pollard_Rho(N); Q=N/P;
    r=(P-1)*(Q-1);
    long long Inv=GetInv(e,r);
    printf("%lld %lld",Inv,Quick_Pow(c,Inv,N));
    return 0;
}