hdu 1211 逆元

RSA

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2353    Accepted Submission(s): 1677

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）

 

 

Source

杭电ACM省赛集训队选拔赛之热身赛

 

 

解密方式 M=c^d mod n

c就是题目中给的数字，n=p*q,现在关键是求d，由  d*e≡1 (mod fn)  fn=(p-1)*(q-1),gcd(e,fn)=1,易得d=e^-1,只要求出e  mod fn的逆元即可。

此逆元不能取模fn又很大，所以费马小定理不是很实用，这里采用拓展欧几里得求逆元。

#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define lld long long
void exGcd (lld a, lld b, lld &d, lld &x, lld &y)
{
    if (b == 0)
    {
        x = 1 ;
        y = 0 ;
        d = a ;
        return ;
    }
    exGcd (b, a%b, d, x, y) ;
    lld tmp = x ;
    x = y ;
    y = tmp - a/b*y ;
}
LL qpow(LL a,LL b,LL c)
{
    LL r=1;
    while(b){
        if(b&1) r=r*a%c;
        a=a*a%c;
        b>>=1;
    }
    return r;
}
int main()
{
    LL p,q,e,n,fn,l;
    lld d,x,y;
    while(cin>>p>>q>>e>>l){n=p*q;
            fn=(p-1)*(q-1);
            exGcd(e,fn,d,x,y);
            x=(x%fn+fn)%fn;
        for(int i=1;i<=l;++i){
            LL num;
            scanf("%lld",&num);
            printf("%c",qpow(num,x,n));
        }cout<<endl;
    }
    return 0;
}