HDU RSA 扩展欧几里得

Problem Description

RSA is one of the most powerful methods to encrypt data. The RSA algorithm is described as follow:

> choose two large prime integer p, q
> 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

You can encrypt data with this method :

C = E(m) = m^e mod n

When you want to decrypt data, use this method :

M = D(c) = c^d mod n

Here, c is an integer ASCII value of a letter of cryptograph and m is an integer ASCII value of a letter of plain text.

Now given p, q, e and some cryptograph, your task is to "translate" the cryptograph into plain text.

 

Input

Each case will begin with four integers p, q, e, l followed by a line of cryptograph. The integers p, q, e, l will be in the range of 32-bit integer. The cryptograph consists of l integers separated by blanks. 

 

Output

For each case, output the plain text in a single line. You may assume that the correct result of plain text are visual ASCII letters, you should output them as visualable letters with no blank between them.

 

Sample Input

101 103 7 11 7716 7746 7497 126 8486 4708 7746 623 7298 7357 3239

 

Sample Output

I-LOVE-ACM.

 

Author

JGShining（极光炫影）

 

#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("
");
    }
}