HDU RSA 扩展欧几里得

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long LL;
/*
题目要求
> calculate n = p × q, calculate F(n) = (p - 1) × (q - 1)
> choose an integer e(1 < e < F(n)), making gcd(e, F(n)) = 1, e will be the public key
> calculate d, making d × e mod F(n) = 1 mod F(n), and d will be the private key
知道 P Q E
gcd(a,b) == 1  等价于 存在x，y  a*x+b*y==1
存在x,y 
e*x + F(n)*y = 1 
d*e%F(n) = 1%F(n)
d*e + F(n)*y = 1; 通过求逆元方法解出 d 即可
*/
LL p,q,e,l;
LL ex_gcd(LL a,LL b,LL &x,LL &y)
{
    if(b==0)
    {
        x = 1;
        y = 0;
        return a;
    }
    LL ans = ex_gcd(b,a%b,x,y);
    LL tmp = x;
    x = y;
    y = tmp - a/b*x;
    return ans;
}
LL cal(LL a,LL b,LL c)
{
    LL x=0,y=0;
    LL gcd = ex_gcd(a,b,x,y);
    if(c%gcd!=0) return -1;
    x *= c/gcd;
    b /= gcd;
    if(b<0) b = -b;
    LL ans = x%b;
    if(ans<0) ans+=b;
    return ans;
}
int main()
{
    while(scanf("%lld%lld%lld%lld",&p,&q,&e,&l)!=EOF)
    {
        LL fn = (p-1)*(q-1),n = p*q;
        LL d = cal(e,fn,1);
        LL tmp,ans;
        for(int i=0;i<l;i++)
        {
            scanf("%lld",&tmp);
            tmp %= n;
            ans = 1;
            for(int j=0;j<d;j++)
                ans = (ans*tmp)%n;
            printf("%c",ans%n);
        }
        printf("\n");
    }
}