2019蓝桥 RSA 解密

#include
#define ll long long 
using namespace std;

inline ll ksc(ll x,ll y,ll mod)
{
    return (x*y-(ll)((long double)x/mod*y)*mod+mod)%mod;     
}

ll fast_pow(ll x, ll k, ll p){
    ll ret=1;
    x%=p;
    while(k>0){
        if(k&1)ret= ksc(ret, x, p);       
        k>>=1;
        x= ksc(x, x, p);
    }
    return ret;
}
ll phi(ll n){
    ll ret= n;
    for(int i=2;i*i<=n;i++){
        if(n%i==0){
            ret= ret/i*(i-1);
            while(n%i==0) n/=i;
        }
    }
    if(n!=1){
        ret= ret/n*(n-1);
    }
    return ret;
}
ll get_p(ll n){
    for(ll i=2;i<=n;i++){
        if(n%i==0){
            return i;
        }
    }
}
int main(){
    ll n = (ll)1001733993063167141;
    ll d = 212353;
    ll C = 20190324;
    ll p,q,e,k;
    printf("n=%lld\n",n);
    p=get_p(n);
    q=n/p;
    printf("p=%lld, q=%lld\n",p,q);
    k=(p-1)*(q-1);
    printf("k=(p-1)*(q-1)=%lld\n",k);
    printf("phi(k)=%lld\n",phi(k));
    e=fast_pow(d,phi(k)-1,k);
    printf("e=d^(phi(k)-1)=%lld (mod k)\n",e);
    printf("d*e=%lld (mod k)\n",ksc(d,e,k));
    ll X=fast_pow(C,e,n);
    printf("X=C^e (mod n)= %lld\n",X);
    return 0;
}

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