【快速幂】HDU1211RSA

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1211

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 117716 7746 7497 126 8486 4708 7746 623 7298 7357 3239

 

Sample Output

I-LOVE-ACM.

代码：

#include<iostream>#include<map>#define LL long longusing namespace std;//  计算a^b%n;快速幂LL power_mod(LL a,LL b,LL n){    LL ans=1;    while(b){        if(b&1)ans=(ans*a)%n;        b>>=1;        a=a*a%n;    }    return ans;}int main(){    int p,q,e,l,c;    cin.sync_with_stdio(false);    while(cin>>p>>q>>e>>l){        LL n=q*p;        map<LL,char>a;      //  开不了这么大的数组,选择用map;        for(int i=0;i<255;i++){            a[power_mod(i,e,n)]=(char)i;        }        for(int i=0;i<l;i++){            cin>>c;            cout<<a[c];        }        cout<<endl;    }    return 0;}/*    通过给定我们密文，我们不易获得明文，    但是给我明文，我们可以很轻松的获得密文；    ASCII总共255个，预处理就可以了*/