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
三品吉他手会点灯2 小时前
C语言学习笔记 - 20.C编程预备计算机专业知识 - 变量为什么必须的初始化【重点】
c语言·笔记·学习
sakiko_2 小时前
UIKit学习笔记1-创建项目(使用UIKit)、使用组件
笔记·学习
生信碱移2 小时前
PACells:这个方法可以鉴定疾病/预后相关的重要细胞亚群,作者提供的代码流程可以学习起来了,甚至兼容转录组与 ATAC 两种数据类型!
人工智能·学习·算法·机器学习·数据挖掘·数据分析·r语言
星幻元宇VR5 小时前
VR航空航天科普设备【VR时空直升机】
科技·学习·安全·生活·vr
_李小白5 小时前
【android opencv学习笔记】Day 2: Mat类(图片数据结构体)
android·opencv·学习
harder3216 小时前
RMP模式的创新突破
开发语言·学习·ios·swift·策略模式
程序猿乐锅7 小时前
【Tilas|第三篇】多表SQL语句
数据库·经验分享·笔记·学习·mysql
徐某人..7 小时前
基于i.MX6ULL平台的智能网关系统开发
arm开发·c++·单片机·qt·物联网·学习·arm
AOwhisky7 小时前
Kubernetes 学习笔记:集群管理、命名空间与 Pod 基础
linux·运维·笔记·学习·云原生·kubernetes
光影少年8 小时前
大屏页面,一次多个请求,请求加密导致 点击 全局时间选择器 时出现卡顿咋解决(面板收起会延迟1~2秒)
前端·javascript·vue.js·学习·前端框架·echarts·reactjs