RSA hdu 1211 数论的水题（数据较弱）

最新推荐文章于 2018-06-15 12:45:00 发布

青苹之末

最新推荐文章于 2018-06-15 12:45:00 发布

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本文详细解析了RSA加密算法的核心原理，并通过代码示例展示了如何使用该算法进行解密操作。通过实例分析，读者可以深入了解RSA算法在实际应用中的解密流程。

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RSA

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1138 Accepted Submission(s): 836

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

Recommend

Eddy

这个题目的真的是相当水，我本来一开始正在发愁又要解线性同余方程了，可是当我看到别人的博客此题数据极

弱，让我颇有点不太适应的，我顿时觉得还是poj的数论题目的比较刺激啊，唉，Romance however, is rocky this

month!!!!感情不顺，只好做数论题目让我开心点！！！

n=p*q, f(n)=(p-1)*(q-1), d*e%f(n)=1, m=c^d%n,我们竟然按照这个式子模拟就能过了！！！！

让我等屌丝惊呆了！！！！！！！！！！！！！！！

#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
    int i,j,p,q,f,e,l,c,m,n,d;
    while(scanf("%d%d%d%d",&p,&q,&e,&l)!=EOF)
    {

      n=p*q;
      f=(p-1)*(q-1);
      d=1;
      while(d*e%f!=1)
      d++;
      for(i=0;i<l;i++)
      {
          scanf("%d",&c);
          int t=1;
          for(j=0;j<d;j++)
          {
              t=(t*c)%n;
          }
          printf("%c",t);
      }
      cout<<endl;
    }
    return 0;
}