zoukankan      html  css  js  c++  java
  • 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) = me mod n

    When you want to decrypt data, use this method :

    M = D(c) = cd 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("
    ");
        }
    }
  • 相关阅读:
    杀死初创科技公司的四大工程陷阱
    杀死初创科技公司的四大工程陷阱
    Linux中su和sudo的用法整理
    Linux中su和sudo的用法整理
    Docket 使用命令
    Docker 部署 portainer
    Linux 双网卡绑定
    docker安装部署,阿里源加速
    nmcli详解
    搭建LAMP环境示例
  • 原文地址:https://www.cnblogs.com/joeylee97/p/6658240.html
Copyright © 2011-2022 走看看