RSA(note)应用密码学第二次作业(9.25)

Basic steps of RSA algorithm

  1. The steps to generate the public key PK and private key SK are as

    follows:

  2. Randomly choose two large prime numbers P and Q. P is not equal to

    Q.

  3. Multiply the two prime numbers P and Q to get an N, that is, N=PQ

  4. Subtract one from P and Q respectively, and then multiply them

    together to get a number T, that is, T = (Q-1) * (P-1)

  5. Choose an integer E as a key so that E and T are relatively prime (

    the greatest common divisor of E and T is 1), and E must be less

    than T

  6. According to the formula DE mod T = 1, the value of D is calculated

    as another key

  7. Through the above steps, the three data N, E, and D can be found,

    where (N, E) is the public key and (N, D) is the private key.

Encrypt information with public key

Plain text: M

Encryption:

Cipher text: C

Decrypt message with private key

Cipher text: C

Decryption:

Plain text: M

In a public-key system using RSA, you intercept the cipher text C = 10 sent lo a user whose public key is e = 5,n = 35. What is the plain text M

answer:

We know that 35 is relatively small, and we can decompose the prime numbers into 5 and 7

Then we can get T=(5-1)X(7-1)=24,24 and 5 are relatively prime

We can use the formula DE mod T=1 to get the value of D which is the private key

We can know the private key from the picture (n,D)=(35,5)

Decrypt

M=5,so the plain text is 5

cpp 复制代码
#include<iostream>
int candp(int b,int p, int k) //b--明文或密文   p--指数(e/d)    k--模数
{
	if (p == 1)
	{
		return b % k;
	}
	if (p == 2)
	{
		return b * b % k;
	}
	if (p % 2 == 0)
	{
		int  sum = candp(b, p / 2, k);
		return sum * sum % k;
	}
	if (p % 2 == 1)
	{
		int sun = candp(b, p / 2, k);
		return sun * sun * b % k;
	}
}
int main()
{
	int d = 1;
	//求e的乘法逆
	int e = 5;
	int t = 24;
	while (((e * d) % t) != 1)
	{
		d++;
	}
	std::cout <<d<<std::endl;


	
		int  n=35, x=0, y=10;

		x = candp(y, d, n);
		std::cout << "明文为:" << x << std::endl;
		return 0;

}
相关推荐
好好研究1 小时前
学习栈和队列的插入和删除操作
数据结构·学习
新中地GIS开发老师2 小时前
新发布:26考研院校和专业大纲
学习·考研·arcgis·大学生·遥感·gis开发·地理信息科学
SH11HF3 小时前
小菜狗的云计算之旅,学习了解rsync+sersync实现数据实时同步(详细操作步骤)
学习·云计算
Frank学习路上3 小时前
【IOS】XCode创建firstapp并运行(成为IOS开发者)
开发语言·学习·ios·cocoa·xcode
Chef_Chen4 小时前
从0开始学习计算机视觉--Day07--神经网络
神经网络·学习·计算机视觉
X_StarX6 小时前
【Unity笔记02】订阅事件-自动开门
笔记·学习·unity·游戏引擎·游戏开发·大学生
MingYue_SSS6 小时前
开关电源抄板学习
经验分享·笔记·嵌入式硬件·学习
weixin_437398217 小时前
转Go学习笔记(2)进阶
服务器·笔记·后端·学习·架构·golang
慕y2747 小时前
Java学习第十六部分——JUnit框架
java·开发语言·学习
peace..8 小时前
温湿度变送器与电脑进行485通讯连接并显示在触摸屏中(mcgs)
经验分享·学习·其他