zoukankan      html  css  js  c++  java
  • HDU1211 RSA

    RSA

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 1415    Accepted Submission(s): 1017


    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.

     
    1和l傻傻分不清。

    //hdu1211
    #include <stdio.h>
    typedef __int64 lld;
    
    lld f(lld c, lld d, lld n)
    {
        if(d == 0) return 1 % n;
        if(d == 1) return c % n;
        lld tmp = f(c, d >> 1, n);
        tmp = tmp * tmp % n;
        if(d & 1) tmp = tmp * c % n;
        return tmp;
    }
    
    int main()
    {
        lld p, q, e, l, c, n, fn, d, i;
        char ch;
        while(scanf("%I64d%I64d%I64d%I64d", &p, &q, &e, &l) != EOF){
            n = p * q; fn = (p - 1) * (q - 1);
            for(i = 1; ; ++i)
                if(i * e % fn == 1) break;
            d = i;
            while(l--){
                scanf("%I64d", &c);
                printf("%c", f(c, d, n));
            }
            printf("
    ");
        }
        return 0;
    }


  • 相关阅读:
    SQL server 日期格式转换style 对应码
    postman的使用方法详解!最全面的教程
    港澳台身份证小结
    使用设置自定义对话框的大小,位置,样式以及设置在安卓桌面上弹出对话框
    android自定义Activity窗口大小(theme运用)
    C#调用RabbitMQ实现消息队列
    C# http请求带请求头部分
    Android如何屏蔽home键和recent键
    针对jquery的优化方法,你知道几条
    试图从目录中执行 CGI、ISAPI 或其他可执行程序
  • 原文地址:https://www.cnblogs.com/yjbjingcha/p/8407083.html
Copyright © 2011-2022 走看看