loj6392 「THUPC2018」密码学第三次小作业 / Rsa
还是挺好做的,\((e_1,e_2)=1 \Rightarrow e_1s+e_2t=0\),\(m \equiv m^1 \equiv m^{e_1s+e_2t} \equiv c_1^s c_2^t\)。exgcd求逆元
#include <iostream>
#include <cstdio>
using namespace std;
typedef long long ll;
int T;
ll exgcd(ll a, ll b, ll &x, ll &y){
if(!b){
x = 1;
y = 0;
return a;
}
ll gcd=exgcd(b, a%b, x, y);
ll z=x;
x = y;
y = z - (a / b) * y;
return gcd;
}
ll mul(ll a, ll b, ll p){
ll re=0;
while(b){
if(b&1) re = (re + a) % p;
a = (a + a) % p;
b >>= 1;
}
return re;
}
ll ksm(ll a, ll b, ll p){
ll re=1;
while(b){
if(b&1) re = mul(re, a, p);
a = mul(a, a, p);
b >>= 1;
}
return re;
}
int main(){
cin>>T;
while(T--){
ll c1, c2, e1, e2, s, t, x, y, n;
scanf("%lld %lld %lld %lld %lld", &c1, &c2, &e1, &e2, &n);
exgcd(e1, e2, s, t);
if(t>0){
ll tmp=t/e1;
s += tmp * e2; t -= tmp * e1;
if(t>0){
s += e2;
t -= e1;
}
}
exgcd(c2, n, x, y);
x = (x % n + n) % n;
t *= -1;
printf("%lld\n", mul(ksm(c1,s,n),ksm(x,t,n),n));
}
return 0;
}
拙いものと思えども、
その手に握る其れこそが、
いつか幻想を生んでいく。