洛谷传送门

 

Solution:（非常言简意赅）

首先分解出p,q 用Pollard's 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:

 1 #include<bits/stdc++.h>
 2 #define Rg register
 3 #define go(i,a,b) for(Rg int i=a;i<=b;i++)
 4 #define ll long long
 5 using namespace std;
 6 ll rd()
 7 {
 8   ll x=0,y=1;char c=getchar();
 9   while(c<'0'||c>'9'){if(c=='-')y=-1;c=getchar();}
10   while(c>='0'&&c<='9'){x=(x<<1)+(x<<3)+c-'0';c=getchar();}
11   return x*y;
12 }
13 ll e,n,c,p,d,g,r;
14 ll mul(ll x,ll y,ll z)
15 {
16   ll sm=0;
17   while(y){if(y&1)sm=(sm+x)%z;x=(x<<1)%z;y>>=1;}
18   return sm;
19 }
20 ll gcd(ll x,ll y){return y==0?x:gcd(y,x%y);}
21 ll f(ll x){return (mul(x,x,n)+g)%n;}
22 void Pollard()
23 {
24   ll a,b;g=rand()%n+1;ll t=0;
25   while(!p)
26   {
27     a=b=rand()%n+1,g=rand()%n+1;
28     while(1)
29     {
30       a=f(a);b=f(f(b));if(a==b)break;
31       ll sm=gcd(abs(a-b),n);
32       if(sm>1){p=sm;break;}
33     }
34   }
35 }
36 ll ksm(ll x,ll y)
37 {
38   ll ans=1;
39   while(y)
40   {if(y&1)ans=mul(ans,x,n);x=mul(x,x,n);y>>=1;}
41   return ans;
42 }
43 ll exgcd(ll a,ll b,ll &x,ll &y)
44 {
45     if(!b){x=1;y=0;return a;}
46     ll d=exgcd(b,a%b,y,x);
47     y-=a/b*x;return d;
48 }
49 int main()
50 {
51   srand(time(0));
52   e=rd(),n=rd(),c=rd();Pollard();
53   r=(p-1)*(n/p-1);exgcd(e,r,d,g);
54   while(d<0)d+=r;while(d>r)d-=r;
55   printf("%lld %lld",d,ksm(c,d));
56   return 0;
57 }

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