(CCPC-Final 2018)K - Mr. Panda and Kakin

题意:x是\([1e5,1e9]\)的随机数,p是小于x的最大素数,q是大于等于x的最小素数,\(n=pq\),\(c=f^{2^{30}+3}\mod{n}\),给n和c求f

题解:rsa解密,首先在\(sqrt(n)\)附近找到p和q,让\(r=(p-1)*(q-1)\),\(e=2^{30}+3\),\(d*e\mod{r}=1\),\(c^d\mod{n}=f\)

证明:\(c=f^e%n\),\(f^{d*e}=f^{d*e\mod(\phi(n))}\mod{n}=f^{d*e\mod{(p-1)*(q-1)}}\mod n=f\mod{n}\)

需要O(1)快速乘

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000009
#define ld long double
//#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
#define ull unsigned long long
//#define base 1000000000000000000
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

inline void add(ll &a,ll b,ll c){a+=b;if(a>=c)a-=c;}
inline ll mul(ll a,ll b,ll c){return (a*b-(ll)((ld)a*b/c)*c+c)%c;}
inline ll qm(ll a,ll b,ll c){ll ans=0;while(b){if(b&1)add(ans,a,c);add(a,a,c);b>>=1;}return ans;}
inline ll qpow(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=mul(ans,a,c);a=mul(a,a,c);b>>=1;}return ans;}

using namespace std;

const ull ba=233;
const db eps=1e-10;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=500000+10,maxn=100000+10,inf=0x3f3f3f3f;

inline ll exgcd(ll a,ll b,ll &x,ll &y)
{
    if(!b){x=1,y=0;return a;}
    ll ans=exgcd(b,a%b,x,y);
    ll t=x;x=y;y=t-a/b*y;
    return ans;
}
int main()
{
    ll a=(1ll<<30)+3;
    int T,cas=0;scanf("%d",&T);
    while(T--)
    {
        ll n,c;scanf("%lld%lld",&n,&c);
        ll p=sqrt(n),q;
        while(n%p!=0)p--;
        q=n/p;
        ll b=(p-1)*(q-1),x,y;
        exgcd(a,b,x,y);
        x=(x%b+b)%b;
        printf("Case %d: %lld\n",++cas,qpow(c,x,n));
    }
    return 0;
}
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