信息计算2019级2班19号程志成

程志成 · 发表于 2022-4-29 22:01:17

第一次

程志成 · 发表于 2022-4-29 22:01:39

第二次.

程志成 · 发表于 2022-4-29 22:01:55

第三次.

程志成 · 发表于 2022-5-12 17:04:50

第四次.

程志成 · 发表于 2022-5-12 17:09:48

import java.math.BigInteger;
import java.util.Random;
import java.util.Scanner;
//RSA算法
public class TestRSA {
@SuppressWarnings("resource")
public static void main(String[] args) {
TestRSA rsa=new TestRSA();
BigInteger[] primeNumber = rsa.getPrimeNumber();
//step1:产生两个大素数
BigInteger p=primeNumber[0];
BigInteger q=primeNumber[1];
System.out.println("1、B产生的两个大素数是(保密)：p="+p+" q="+q);
//step2.1:计算n
BigInteger n=p.multiply(q);//n=p*q
System.out.println("2、计算的n是："+n);
//step2.2:计算sn
BigInteger sn=(p.subtract(new BigInteger("1"))).multiply(q.subtract(new BigInteger("1")));//sn=(p-1)*(q-1)
System.out.println(" 计算的sn是："+sn);
//step3:随机选取e
BigInteger e=rsa.getE(sn);//0<e<sn && e和 sn互素
System.out.println("3、选取的e是："+e);
//step4:计算d
BigInteger d=rsa.getD(sn, e).mod(sn);//d同时与n和sn互素
System.out.println("4、计算的d是："+d);
//step5:B将n和e作为公钥公开
System.out.println("5、公钥：n="+n+" e="+e);
//step6:用户A获取到公钥
System.out.println("6、用户A已经获取到公钥了，可以开始发数据了...");
//step7:进行加密
System.out.println("7、请输入要发送的明文（仅数字）:");
Scanner scanner=new Scanner(System.in);
BigInteger m=new BigInteger(scanner.next());
BigInteger c=rsa.getC(m,e,n);//计算密文
System.out.println("加密后的密文是："+c);
//step8:进行解密
BigInteger mm=rsa.getDecrypt(c,n,d);//进行解密
System.out.println("8、解密后的的结果是："+mm);
}
public BigInteger[] getPrimeNumber(){
BigInteger p=null;
BigInteger q=null;
BigInteger[] res=new BigInteger[2];
Random random = new Random();
p=BigInteger.probablePrime(64, random);
q=BigInteger.probablePrime(64, random);
res[0]=p;
res[1]=q;
return res;
}
public BigInteger getE(BigInteger sn){
BigInteger e = null;
int length = sn.toString().length()-2;
e=new BigInteger(sn.toString().subSequence(0, length-2).toString()).nextProbablePrime();
return e;
}
public BigInteger getD(BigInteger sn,BigInteger e){
BigInteger[] ret = new BigInteger[3];
BigInteger u = BigInteger.valueOf(1), u1 = BigInteger.valueOf(0);
BigInteger v = BigInteger.valueOf(0), v1 = BigInteger.valueOf(1);
if (e.compareTo(sn) > 0) {
BigInteger tem = sn;
sn = e;
e = tem;
}
while (e.compareTo(BigInteger.valueOf(0)) != 0) {
BigInteger tq = sn.divide(e);
BigInteger tu = u;
u = u1;
u1 = tu.subtract(tq.multiply(u1));
BigInteger tv = v;
v = v1;
v1 = tv.subtract(tq.multiply(v1));
BigInteger tsn = sn;
sn = e;
e = tsn.subtract(tq.multiply(e));
ret[0] = u;
ret[1] = v;
ret[2] = sn;
}
return ret[1];
}
public BigInteger getC(BigInteger m,BigInteger e,BigInteger n){
BigInteger c=null;
c=m.modPow(e, n);
return c;
}
public BigInteger getDecrypt(BigInteger c,BigInteger n,BigInteger d){
BigInteger m=null;
m=c.modPow(d, n);
return m;
}
}

复制代码

程志成 · 发表于 2022-6-2 18:26:04

ecc算法

#include <stdio.h>
#include <math.h>
int xy[22];
#define N 23
#define A 1
//#define M 29
#define a1 1
#define b1 4
typedef struct point {
int x;
int y;
}Point;
typedef struct ecc {
struct point p[100];
int len;
}ECCPoint;
typedef struct generator {
Point p;
int p_class;
}GENE_SET;
ECCPoint eccPoint;
Point mul(Point p1, Point p2);
Point add_two_points(Point p1, Point p2);
GENE_SET geneSet[100];
int geneLen;
//判断平方根是否为整数
int int_sqrt(int s)
{
int temp;
temp = (int)sqrt(s);//转为整型
if (temp * temp == s)
{
return temp;
}
else {
return -1;
}
}
//取模函数
int mod_p(int s)
{
int i; //保存s/p的倍数
int result; //模运算的结果
i = s / N;
result = s - i * N;
if (result >= 0)
{
return result;
}
else
{
return result + N;
}
}
//打印点集
void print()
{
int i;
int len = eccPoint.len;
printf("\n该椭圆曲线上共有%d个点(包含无穷远点)\n", len + 1);
for (i = 0; i < len; i++)
{
if (i % 8 == 0)
{
printf("\n");
}
printf("(%2d,%2d)\t", eccPoint.p[i].x, eccPoint.p[i].y);
}
printf("\n");
}
//task1:求出椭圆曲线上所有点
void get_all_points()
{
int i = 0;
int j = 0;
int s, y = 0;
int n = 0, q = 0;
int modsqrt = 0;
int flag = 0;
if (a1 * a1 * a1 + b1 * b1 + 4 != 0)
{
for (i = 0; i <= N - 1; i++)
{
flag = 0;
n = 1;
y = 0;
s = i * i * i + a1 * i + b1;
while (s < 0)
{
s += N;
}
s = mod_p(s);
modsqrt = int_sqrt(s);
if (modsqrt != -1)
{
flag = 1;
y = modsqrt;
}
else {
while (n <= N - 1)
{
q = s + n * N;
modsqrt = int_sqrt(q);
if (modsqrt != -1)
{
y = modsqrt;
flag = 1;
break;
}
flag = 0;
n++;
}
}
if (flag == 1)
{
eccPoint.p[j].x = i;
eccPoint.p[j].y = y;
j++;
if (y != 0)
{
eccPoint.p[j].x = i;
eccPoint.p[j].y = (N - y) % N;
j++;
}
}
}
eccPoint.len = j;//点集个数
print(); //打印点集
}
}
//task3:求生成元以及阶
void get_generetor_class()
{
int i, j = 0;
int count = 1;
Point p1, p2;
get_all_points();
printf("\n**********************************输出生成元以及阶：*************************************\n");
for (i = 0; i < eccPoint.len; i++)
{
count = 1;
p1.x = p2.x = eccPoint.p[i].x;
p1.y = p2.y = eccPoint.p[i].y;
while (1)
{
p2 = add_two_points(p1, p2);
if (p2.x == -1 && p2.y == -1)
{
break;
}
count++;
if (p2.x == p1.x)
{
break;
}
}
count++;
if (count <= eccPoint.len + 1)
{
geneSet[j].p.x = p1.x;
geneSet[j].p.y = p1.y;
geneSet[j].p_class = count;
printf("(%d,%d)--->>%d\t", geneSet[j].p.x, geneSet[j].p.y, geneSet[j].p_class);
j++;
if (j % 6 == 0) {
printf("\n");
}
}
geneLen = j;
}
}
// 求 a mod b 的逆元
void exGcd(int a, int b) {
if (b == 0) {
xy[0] = 1;
xy[1] = 0;
}
else {
exGcd(b, a % b);
int x = xy[0];
xy[0] = xy[1];
xy[1] = x - (a / b) * xy[1];
}
}
int calculate3(int y, int k, int p) {
int l = 1;
for (int i = 0; i < k; i++) {
l = l * y;
l = l % p;
}
return l;
}
Point eccmutiply(int n, Point p) {
int a, b, l, k, m;
a = p.x;
b = p.y;
for (int i = 0; i < n - 1; i++) {
if (a == p.x && b == p.y) {
exGcd(2 * p.y, N);
k = xy[0];
if (k < 0)k = k + N;
printf("逆元=%d\n", k);
l = (3 * p.x * p.x + A) * k;
l = calculate3(l, 1, N);
if (l < 0) {
l = l + N;
}
}
else {
exGcd(a - p.x + N, N);
k = xy[0];
if (k < 0)k = k + N;
printf("else逆元=%d\n", k);
l = (b - p.y) * k;
l = calculate3(l, 1, N);
if (l < 0) {
l = l + N;
}
printf("l=%d\n", l);
}
m = p.x;
a = l * l - a - p.x;
a = calculate3(a, 1, N);
if (a < 0) {
a = a + N;
}
b = l * (m - a) - p.y;
b = calculate3(b, 1, N);
if (b < 0) {
b = b + N;
}
printf("%d(a,b)=(%d,%d)\n", i + 2, a, b);
//if(a==4&&b==5)break;
}
Point p3;
p3.x = a;
p3.y = b;
return p3;
}
Point mul(Point p1, Point p2) {
int k, l;
exGcd(p2.x - p1.x + N, N);
k = xy[0];
if (k < 0)k = k + N;
//printf("else逆元=%d\n",k);
l = (p2.y - p1.y) * k;
l = calculate3(l, 1, N);
if (l < 0) {
l = l + N;
}
//printf("l=%d\n",l);
Point p3;
p3.x = l * l - p1.x - p2.x;
p3.x = calculate3(p3.x, 1, N);
if (p3.x < 0)p3.x = p3.x + N;
p3.y = l * (p1.x - p3.x) - p1.y;
p3.y = calculate3(p3.y, 1, N);
if (p3.y < 0)p3.y = p3.y + N;
return p3;
}
//求b关于n的逆元
int inverse(int n, int b)
{
int q, r, r1 = n, r2 = b, t, t1 = 0, t2 = 1, i = 1;
while (r2 > 0)
{
q = r1 / r2;
r = r1 % r2;
r1 = r2;
r2 = r;
t = t1 - q * t2;
t1 = t2;
t2 = t;
}
if (t1 >= 0)
return t1 % n;
else {
while ((t1 + i * n) < 0)
i++;
return t1 + i * n;
}
}
//两点的加法运算
Point add_two_points(Point p1, Point p2)
{
long t;
int x1 = p1.x;
int y1 = p1.y;
int x2 = p2.x;
int y2 = p2.y;
int tx, ty;
int x3, y3;
int flag = 0;
//求
if ((x2 == x1) && (y2 == y1))
{
//相同点相加
if (y1 == 0)
{
flag = 1;
}
else {
t = (3 * x1 * x1 + a1) * inverse(N, 2 * y1) % N;
}
//printf("inverse(p,2*y1)=%d\n",inverse(p,2*y1));
}
else {
//不同点相加
ty = y2 - y1;
tx = x2 - x1;
while (ty < 0)
{
ty += N;
}
while (tx < 0)
{
tx += N;
}
if (tx == 0 && ty != 0)
{
flag = 1;
}
else {
t = ty * inverse(N, tx) % N;
}
}
if (flag == 1)
{
p2.x = -1;
p2.y = -1;
}
else {
x3 = (t * t - x1 - x2) % N;
y3 = (t * (x1 - x3) - y1) % N;
//使结果在有限域GF(P)上
while (x3 < 0)
{
x3 += N;
}
while (y3 < 0)
{
y3 += N;
}
p2.x = x3;
p2.y = y3;
}
return p2;
}
//task2:倍点运算的递归算法
Point timesPiont(int k, Point p0)
{
if (k == 1) {
return p0;
}
else if (k == 2) {
return add_two_points(p0, p0);
}
else {
return add_two_points(p0, timesPiont(k - 1, p0));
}
}
main() {
get_generetor_class();
Point p1, p2, p, p4, p5, p6, p7, p8, p9, p10;
int na, k, h, r, s, u1, u2;
printf("选择生成元");
scanf_s("%d%d", &p1.x, &p1.y);
int j = 0;
while (j < N) {
if (geneSet[j].p.x == p1.x && geneSet[j].p.y == p1.y) {
break;
}
printf("j=%d\n", j);
++j;
}
int M = geneSet[j].p_class;
printf("M=%d", M);
// p1.x=0;
// p1.y=2;
//p.x=20;
//p.y=1;
printf("请输入私钥");
scanf_s("%d", &na);
p2 = eccmutiply(na, p1);
printf("公钥为（%d,%d)", p2.x, p2.y);
//p2=mul(p,p1);
//printf("(%d,%d)",p2.x,p2.y);
//签名过程
printf("输入随机数k\n");
scanf_s("%d", &k);
printf("输入hash\n");
scanf_s("%d", &h);
p4 = eccmutiply(k, p1);
r = calculate3(p4.x, 1, N);
if (r < 0)r = r + N;
s = inverse(M, k) * (h + na * r) % M;
if (s < 0)s = s + N;
printf("签名为(%d,%d)\n", r, s);
printf("========================");
//验证过程
u1 = h * inverse(M, s) % M;
if (u1 < 0)u1 = u1 + M;
u2 = r * inverse(M, s) % M;
if (u2 < 0)u2 = u2 + M;
printf("u1u2=%d,%d\n", u1, u2);
p5 = eccmutiply(u1, p1);
printf("sG=(%d,%d)\n", p5.x, p5.y);
p6 = eccmutiply(u2, p2);
printf("HP=(%d,%d)\n", p6.x, p6.y);
p7 = add_two_points(p5, p6);
printf("sG+HP=(%d,%d)\n", p7.x, p7.y);
if (calculate3(p7.x, 1, N) == r) {
printf("通过");
}
else {
printf("失败");
}
}

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信息计算2019级2班19号程志成

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