gym-102055-RSA加密-欧拉降幂

题目链接

gym102055

思路

首先要明白一个模运算的性质
(a ^ b) % p = ((a % p)^b) % p
$令e=2^{30}+3,r=(q-1)(p-1)$, $找到e在模n下的逆元d，de=1(mod\ n)$,然后因为 $c=f^{2^{30}+3}(mod\ n)$,此处用欧拉降幂就得到 $c=f^{d^{-1}}(mod\ n)$,运用上面第四条性质就可以得到 $c^d=f(mod\ n)$因为此处的e与n不一定互质，所以用扩展欧几里得求逆元，然后快速幂求解即可，然后中间用上快速乘就可以过了

代码实现

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define ld long double
const int mod=1e9+7;
inline ll mul(ll a,ll b,ll c){return (a*b-(ll)((ld)a*b/c)*c+c)%c;}
inline ll exgcd(ll a,ll b,ll &x,ll &y){if(!b){x=1;y=0;return a;}ll g = exgcd(b,a%b,y,x);y-=a/b*x;return g;}
inline ll quick_pow(ll a,ll b,ll mod){ll res=1;while(b){if(b&1)res=mul(res,a,mod);a=mul(a,a,mod);b>>=1;}return res;}
inline ll quick_pow(ll a,ll b){ll res=1;while(b){if(b&1)res=mul(res,a,mod);a=mul(a,a,mod);b>>=1;}return res;}
inline ll inv(ll x){return quick_pow(x,mod-2);}
inline ll inv(ll x,ll mod){return quick_pow(x,mod-2,mod);}
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
int main(){
	int t;
	scanf("%d",&t);
	ll a = (1<<30)+3;
	for(int cas=1;cas<=t;cas++){
		ll n,c;
		scanf("%lld %lld",&n,&c);
		ll p = sqrt(n);
		while(n%p!=0)p--;
		ll q = n/p;
		ll r = (p-1)*(q-1);
		ll x,y;
		exgcd(a,r,x,y);
		x=((x%r)+r)%r;
		printf("Case %d: %lld\n",cas,quick_pow(c,x,n));
	}
}