CQOI2016 密钥破解

洛谷传送门

Solution:（非常言简意赅）

首先分解出p,q 用Pollard-Rho算法

算出r

解ed≡1(mod r)的同余方程算出d

用快速幂算出c^dmod N 然后没啦

Code:

#include<bits/stdc++.h>
#define Rg register
#define go(i,a,b) for(Rg int i=a;i<=b;i++)
#define ll long long
using namespace std;
ll rd()
{
  ll x=0,y=1;char c=getchar();
  while(c<'0'||c>'9'){if(c=='-')y=-1;c=getchar();}
  while(c>='0'&&c<='9'){x=(x<<1)+(x<<3)+c-'0';c=getchar();}
  return x*y;
}
ll e,n,c,p,d,g,r;
ll mul(ll x,ll y,ll z)
{
  ll sm=0;
  while(y){if(y&1)sm=(sm+x)%z;x=(x<<1)%z;y>>=1;}
  return sm;
}
ll gcd(ll x,ll y){return y==0?x:gcd(y,x%y);}
ll f(ll x){return (mul(x,x,n)+g)%n;}
void Pollard()
{
  ll a,b;g=rand()%n+1;ll t=0;
  while(!p)
  {
    a=b=rand()%n+1,g=rand()%n+1;
    while(1)
    {
      a=f(a);b=f(f(b));if(a==b)break;
      ll sm=gcd(abs(a-b),n);
      if(sm>1){p=sm;break;}
    }
  }
}
ll ksm(ll x,ll y)
{
  ll ans=1;
  while(y)
  {if(y&1)ans=mul(ans,x,n);x=mul(x,x,n);y>>=1;}
  return ans;
}
ll exgcd(ll a,ll b,ll &x,ll &y)
{
    if(!b){x=1;y=0;return a;}
    ll d=exgcd(b,a%b,y,x);
    y-=a/b*x;return d;
}
int main()
{
  srand(time(0));
  e=rd(),n=rd(),c=rd();Pollard();
  r=(p-1)*(n/p-1);exgcd(e,r,d,g);
  while(d<0)d+=r;while(d>r)d-=r;
  printf("%lld %lld",d,ksm(c,d));
  return 0;
}