洛咕
题意:对于整数N(\(N<=2^{62}\)),求出两个不同的质数p和q,使得\(N=p*q\);令\(r=(p-1)*(q-1)\),给定整数e(e<r且e,r互质),求出整数d,使得\(e*d≡1(\mod r)\);给定整数c,求出整数n使得\(c*d≡n(\mod N)\).
#include<bits/stdc++.h>
#define rg register
#define LL long long
using namespace std;
inline LL read(){
rg LL s=0,w=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){s=s*10+ch-'0';ch=getchar();}
return s*w;
}
LL e,N,c;
inline LL ksc(LL a,LL b,LL c){
rg LL cnt=0;
while(b){
if(b&1)cnt=(cnt+a)%c;
a=(a+a)%c;
b>>=1;
}
return cnt%c;
}//龟速乘,防止乘的时候爆long long
inline LL ksm(LL a,LL b,LL c){
rg LL cnt=1;
while(b){
if(b&1)cnt=ksc(cnt,a,c);
a=ksc(a,a,c);
b>>=1;
}
return cnt%c;
}
inline LL gcd(LL a,LL b){
if(b==0)return a;
return gcd(b,a%b);
}
inline LL pr(LL n){
rg LL x,y,cs,sum,i,j,d;
while(1){
y=x=rand()%n;cs=rand()%n;
sum=1;i=0;j=1;
while(++i){
x=ksc(x,x,n)+cs;
sum=ksc(abs(y-x),sum,n);
if(x==y||sum==0)break;
if(i%127==0||i==j){
d=gcd(sum,n);
if(d>1&&d!=n)return d;
if(i==j)y=x,j<<=1;
}
}
}
}
inline LL exgcd(LL a,LL b,LL &x,LL &y){
if(b==0){x=1;y=0;return a;}
LL d=exgcd(b,a%b,y,x);
y-=(a/b)*x;
return d;
}
int main(){
e=read();N=read();c=read();
LL p=pr(N),r=1LL*(p-1)*(N/p-1);
LL x,y;exgcd(e,r,x,y);
LL d=(x%r+r)%r,n=ksm(c,d,N);
printf("%lld %lld\n",d,n);
return 0;
}