OpenSSL入门-椭圆曲线

数学基础

例如求椭圆曲线$y^2=x^3-x$在有限域$\operatorname{GF}(89)$上所有点：

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#include <iostream>
#include<string.h>
#include<math.h>
#include<time.h>
#define MAX 100
typedef struct point {
    int point_x;
    int point_y;
}Point;
typedef struct ecc {
    struct point p[MAX];
    int len;
}ECCPoint;
typedef struct generator {
    Point p;
    int p_class;
}GENE_SET;
char alphabet[] = "abcdefghijklmnopqrstuvwxyz";
int a = -1, b = 0, p = 89;//椭圆曲线为E89(-1,0)： y2=x3-x (mod 89)
ECCPoint eccPoint;
GENE_SET geneSet[MAX];
int geneLen;
char plain[] = "yes";
int m[MAX];
int cipher[MAX];
int nB;//私钥
Point P1, P2, Pt, G, PB;
Point Pm;
int C[MAX];
//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 / p;
    result = s - i * p;
    if (result >= 0)
        return result;
    else
        return result + p;
}
//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;
}
//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].point_x, eccPoint.p[i].point_y);
    }
    printf("\n");
}
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 (4 * a * a * a + 27 * b * b != 0) {
        for (i = 0; i <= p - 1; i++) {
            flag = 0;
            n = 1;
            y = 0;
            s = i * i * i + a * i + b;
            while (s < 0)
                s += p;
            s = mod_p(s);
            modsqrt = int_sqrt(s);
            if (modsqrt != -1) {
                flag = 1;
                y = modsqrt;
            }
            else
                while (n <= p - 1) {
                    q = s + n * p;
                    modsqrt = int_sqrt(q);
                    if (modsqrt != -1) {
                        y = modsqrt;
                        flag = 1;
                        break;
                    }
                    flag = 0;
                    n++;
                }
            if (flag == 1) {
                eccPoint.p[j].point_x = i;
                eccPoint.p[j].point_y = y;
                j++;
                if (y != 0) {
                    eccPoint.p[j].point_x = i;
                    eccPoint.p[j].point_y = (p - y) % p;  //注意：P(x,y)的负元是 (x,p-y)
                    j++;
                }
            }
        }
        eccPoint.len = j;//点集个数 
        print(); //打印点集 
    }
}
int main() {
    get_all_points();
}

例如实现点加和倍点算法，其中点加运算公式为：

设$P(x_1,y_1),Q(x_2,y_2)$，得$P+Q=R(x_3,y_3)$，有：
$$
\begin{cases}
x_3=\left(\dfrac{y_2-y_1}{x_2-x_1}\right)^2-x_1-x_2,\\
y_3=-\left(\dfrac{y_2-y_1}{x_2-x_1}\right)x_3+\left(\dfrac{y_2-y_1}{x_2-x_1}\right)x_1+y_1
\end{cases}
$$
对于倍点运算，$P+Q=2P=R$，有：
$$
\begin{cases}
x_3=\left(\dfrac{3x_1^2+a}{2y_1}\right)^2-2x_1,\\
y_3=-\left(\dfrac{3x_1^2+a}{2y_1}\right)x_3+\left(\dfrac{3x_1^2+a}{2y_1}\right)x_1-y_1
\end{cases}
$$
实现为：

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#include <iostream>
#define MAX 100
typedef struct point {
    int point_x;
    int point_y;
}Point;
int a = -1, b = 0, p = 89;//椭圆曲线为E89(-1,0)： y2=x3-x (mod 89)
//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.point_x;
    int y1 = p1.point_y;
    int x2 = p2.point_x;
    int y2 = p2.point_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 + a) * inverse(p, 2 * y1) % p;
        //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 += p;
        while (tx < 0)
            tx += p;
        if (tx == 0 && ty != 0)
            flag = 1;
        else
            t = ty * inverse(p, tx) % p;
    }
    if (flag == 1) {
        p2.point_x = -1;
        p2.point_y = -1;
    }
    else {
        x3 = (t * t - x1 - x2) % p;
        y3 = (t * (x1 - x3) - y1) % p;
        //使结果在有限域GF(P)上 
        while (x3 < 0)
            x3 += p;
        while (y3 < 0)
            y3 += p;
        p2.point_x = x3;
        p2.point_y = y3;
    }
    return p2;
}
int main() {
    Point p1, p2, p3;
    p1.point_x = 6; p1.point_y = 78;
    p2.point_x = 7; p2.point_y = 43;
    p3 = add_two_points(p1, p2);
    printf("p3=(%d,%d)\n", p3.point_x, p3.point_y);
}

标量乘运算时，一旦相加得两个点相同就是倍点运算，不同就是点加运算。

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#include <iostream>
#define MAX 100
typedef struct point {
    int point_x;
    int point_y;
}Point;
typedef struct ecc {
    struct point p[MAX];
    int len;
}ECCPoint;
int a = -1, b = 0, p = 89;//椭圆曲线为E89(-1,0)： y2=x3-x (mod 89)
//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;
    }
}
Point add_two_points(Point p1, Point p2) {
    long t;
    int x1 = p1.point_x;
    int y1 = p1.point_y;
    int x2 = p2.point_x;
    int y2 = p2.point_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 + a) * inverse(p, 2 * y1) % p;
        //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 += p;
        while (tx < 0)
            tx += p;
        if (tx == 0 && ty != 0)
            flag = 1;
        else
            t = ty * inverse(p, tx) % p;
    }
    if (flag == 1) {
        p2.point_x = -1;
        p2.point_y = -1;
    }
    else {
        x3 = (t * t - x1 - x2) % p;
        y3 = (t * (x1 - x3) - y1) % p;
        //使结果在有限域GF(P)上 
        while (x3 < 0)
            x3 += p;
        while (y3 < 0)
            y3 += p;
        p2.point_x = x3;
        p2.point_y = y3;
    }
    return p2;
}
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));
}
int main() {
    Point pt;
    pt.point_x = 6; pt.point_y = 78;
    Point p = timesPiont(5, pt);
    printf("p=(%d,%d)\n", p.point_x, p.point_y);
}

若椭圆曲线上一点$P$存在最小正整数$n$使得数乘$nP=O\infin$，则称$n$为$P$的阶。这时对点$P$不停点乘运算，当某次点上横坐标等于$P$的横坐标时，次数$n+1$则为阶。例如计算$y^2=x^3-x$的所有阶。

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#include <iostream>
#define MAX 100
typedef struct point {
    int point_x;
    int point_y;
}Point;
typedef struct ecc {
    struct point p[MAX];
    int len;
}ECCPoint;
typedef struct generator {
    Point p;
    int p_class;
}GENE_SET;
ECCPoint eccPoint;
GENE_SET geneSet[MAX];
int geneLen;
int a = -1, b = 0, p = 89;//椭圆曲线为E89(-1,0)： y2=x3-x (mod 89)
//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].point_x, eccPoint.p[i].point_y);
    }
    printf("\n");
}
//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 / p;
    result = s - i * p;
    if (result >= 0)
        return result;
    else
        return result + p;
}
//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;
}
//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;
    }
}
//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 (4 * a * a * a + 27 * b * b != 0) {
        for (i = 0; i <= p - 1; i++) {
            flag = 0;
            n = 1;
            y = 0;
            s = i * i * i + a * i + b;
            while (s < 0)
                s += p;
            s = mod_p(s);
            modsqrt = int_sqrt(s);
            if (modsqrt != -1) {
                flag = 1;
                y = modsqrt;
            }
            else
                while (n <= p - 1) {
                    q = s + n * p;
                    modsqrt = int_sqrt(q);
                    if (modsqrt != -1) {
                        y = modsqrt;
                        flag = 1;
                        break;
                    }
                    flag = 0;
                    n++;
                }
            if (flag == 1) {
                eccPoint.p[j].point_x = i;
                eccPoint.p[j].point_y = y;
                j++;
                if (y != 0) {
                    eccPoint.p[j].point_x = i;
                    eccPoint.p[j].point_y = (p - y) % p;
                    j++;
                }
            }
        }
        eccPoint.len = j;//点集个数 
        print(); //打印点集 
    }
}
//两点的加法运算 
Point add_two_points(Point p1, Point p2) {
    long t;
    int x1 = p1.point_x;
    int y1 = p1.point_y;
    int x2 = p2.point_x;
    int y2 = p2.point_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 + a) * inverse(p, 2 * y1) % p;
        //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 += p;
        while (tx < 0)
            tx += p;
        if (tx == 0 && ty != 0)
            flag = 1;
        else
            t = ty * inverse(p, tx) % p;
    }
    if (flag == 1) {
        p2.point_x = -1;
        p2.point_y = -1;
    }
    else {
        x3 = (t * t - x1 - x2) % p;
        y3 = (t * (x1 - x3) - y1) % p;
        //使结果在有限域GF(P)上 
        while (x3 < 0)
            x3 += p;
        while (y3 < 0)
            y3 += p;
        p2.point_x = x3;
        p2.point_y = y3;
    }
    return p2;
}
//求点的阶
void get_generetor_class() {
    int i, j = 0;
    int count = 1;
    Point p1, p2;
    get_all_points();
    printf("\n*************输出所有点的阶：*******************\n");
    for (i = 0; i < eccPoint.len; i++) {
        count = 1;
        p1.point_x = p2.point_x = eccPoint.p[i].point_x;
        p1.point_y = p2.point_y = eccPoint.p[i].point_y;
        while (1) {
            p2 = add_two_points(p1, p2);
            if (p2.point_x == -1 && p2.point_y == -1)
                break;
            count++;
            if (p2.point_x == p1.point_x)
                break;
        }
        count++;
        if (count <= eccPoint.len + 1) {
            geneSet[j].p.point_x = p1.point_x;
            geneSet[j].p.point_y = p1.point_y;
            geneSet[j].p_class = count;
            printf("点(%d,%d)的阶：%d\t", geneSet[j].p.point_x, geneSet[j].p.point_y, geneSet[j].p_class);
            j++;
            if (j % 3 == 0)
                printf("\n");
        }
        geneLen = j;
    }
}
int main() {
    get_generetor_class();
    return 0;
}

ECC

考虑$K=kG$，$K,G$为$E_p(a,b)$上的点，$nG=O\infty,n\in\mathbb P$，$k<n,k\in\mathbb Z$。其中$G$称为基点或生成元，$k$为私钥，$K$为公钥。加解密通信过程如下：

  • A选定一条$E_p(a,b)$，并取一基点$G$。
  • A选一个私钥$k$，生成公钥$K=kG$。
  • A将$E_p(a,b)$和点$K,G$传给B。
  • B将待传输的明文编码到$E_p(a,b)$上一点$M$，产生一个随机整数$r$，计算点$C_1=M+r,C_2=rG$，将$C_1,C_2$传给A。
  • A计算$C_1-kC_2$，结果为点$M$。

中间人通过$K,G$求$k$或通过$C_2,G$求$r$都困难。

描述一条$F_p$上椭圆曲线用$T=(p,a,b,G,n,h)$。$G$为基点，$n$为基点的阶，$m$为曲线所有点个数，$h=\left\lfloor\dfrac{m}{n}\right\rfloor$，一般要求条件如下：

  • $p$越大越好，一般200位左右。
  • $p\neq nh$。
  • $p^t\not\equiv1\pmod n,1\leqslant t<20$。
  • $4a^3+27b^2\not\equiv0\pmod p$。
  • $h\leqslant4$。

实现如下：

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#include <iostream>
#include<math.h>
#include<time.h>
#define MAX 100
typedef struct point {
    int point_x;
    int point_y;
}Point;
typedef struct ecc {
    struct point p[MAX];
    int len;
}ECCPoint;
typedef struct generator {
    Point p;
    int p_class;
}GENE_SET;
char alphabet[] = "abcdefghijklmnopqrstuvwxyz";
int a = -1, b = 0, p = 89;//椭圆曲线为E89(-1,0)： y2=x3-x (mod 89)
ECCPoint eccPoint;
GENE_SET geneSet[MAX];
int geneLen;
char plain[] = "yes";
int m[MAX];
int cipher[MAX];
int nB;//私钥 
Point P1, P2, Pt, G, PB;
Point Pm;
int C[MAX];
//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].point_x, eccPoint.p[i].point_y);
    }
    printf("\n");
}
//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 / p;
    result = s - i * p;
    if (result >= 0)
        return result;
    else
        return result + p;
}
//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;
}
//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;
    }
}
//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 (4 * a * a * a + 27 * b * b != 0){
        for (i = 0; i <= p - 1; i++){
            flag = 0;
            n = 1;
            y = 0;
            s = i * i * i + a * i + b;
            while (s < 0)
                s += p;
            s = mod_p(s);
            modsqrt = int_sqrt(s);
            if (modsqrt != -1){
                flag = 1;
                y = modsqrt;
            }
            else
                while (n <= p - 1){
                    q = s + n * p;
                    modsqrt = int_sqrt(q);
                    if (modsqrt != -1){
                        y = modsqrt;
                        flag = 1;
                        break;
                    }
                    flag = 0;
                    n++;
                }
            if (flag == 1){
                eccPoint.p[j].point_x = i;
                eccPoint.p[j].point_y = y;
                j++;
                if (y != 0){
                    eccPoint.p[j].point_x = i;
                    eccPoint.p[j].point_y = (p - y) % p;
                    j++;
                }
            }
        }
        eccPoint.len = j;//点集个数 
        print(); //打印点集 
    }
}
//两点的加法运算 
Point add_two_points(Point p1, Point p2){
    long t;
    int x1 = p1.point_x;
    int y1 = p1.point_y;
    int x2 = p2.point_x;
    int y2 = p2.point_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 + a) * inverse(p, 2 * y1) % p;
        //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 += p;
        while (tx < 0)
            tx += p;
        if (tx == 0 && ty != 0)
            flag = 1;
        else
            t = ty * inverse(p, tx) % p;
    }
    if (flag == 1){
        p2.point_x = -1;
        p2.point_y = -1;
    }
    else {
        x3 = (t * t - x1 - x2) % p;
        y3 = (t * (x1 - x3) - y1) % p;
        //使结果在有限域GF(P)上 
        while (x3 < 0)
            x3 += p;
        while (y3 < 0)
            y3 += p;
        p2.point_x = x3;
        p2.point_y = y3;
    }
    return p2;
}
//求点的阶
void get_generetor_class(){
    int i, j = 0;
    int count = 1;
    Point p1, p2;
    get_all_points();
    printf("\n*************输出所有点的阶：*******************\n");
    for (i = 0; i < eccPoint.len; i++){
        count = 1;
        p1.point_x = p2.point_x = eccPoint.p[i].point_x;
        p1.point_y = p2.point_y = eccPoint.p[i].point_y;
        while (1){
            p2 = add_two_points(p1, p2);
            if (p2.point_x == -1 && p2.point_y == -1)
                break;
            count++;
            if (p2.point_x == p1.point_x)
                break;
        }
        count++;
        if (count <= eccPoint.len + 1){
            geneSet[j].p.point_x = p1.point_x;
            geneSet[j].p.point_y = p1.point_y;
            geneSet[j].p_class = count;
            printf("点(%d,%d)的阶：%d\t", geneSet[j].p.point_x, geneSet[j].p.point_y, geneSet[j].p_class);
            j++;
            if (j % 3 == 0)
                printf("\n");
        }
        geneLen = j;
    }
}
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));
}
//判断是否为素数 
int isPrime(int n){
    int i, k;
    k = sqrt(n);
    for (i = 2; i <= k; i++)
        if (n % i == 0)
            break;
    if (i <= k)
        return -1;
    else
        return 0;
}
//加密 
void encrypt_ecc(){
    int num, i, j;
    int gene_class;
    int num_t;
    int k;
    printf("\n\n明文数据：%s\n", plain);
    srand(time(NULL));
    //明文转换过程
    for (i = 0; i < strlen(plain); i++)
        for (j = 0; j < 26; j++) //for(j=0;j<26;j++) 
            if (plain[i] == alphabet[j])
                m[i] = j;//将字符串明文换成数字，并存到整型数组m里面
    //选择生成元
    num = rand() % geneLen;
    gene_class = geneSet[num].p_class;
    while (isPrime(gene_class) == -1) {//不是素数 
        num = rand() % (geneLen - 3) + 3;
        gene_class = geneSet[num].p_class;
    }
    //printf("gene_class=%d\n", gene_class);
    //printf("gene_class=%d\n",gene_class);
    G = geneSet[num].p;
    //printf("G:(%d,%d)\n",geneSet[num].p.point_x,geneSet[num].p.point_y);
    nB = rand() % (gene_class - 1) + 1;//选择私钥 
    PB = timesPiont(nB, G);//PB是公钥 
    printf("\n公钥：\n");
    printf("{y^2=x^3%d*x+%d,%d,(%d,%d),(%d,%d)}\n", a, b, gene_class, G.point_x, G.point_y, PB.point_x, PB.point_y);
    printf("私钥：\n");
    printf("nB=%d\n", nB);
    //加密 
    k = rand() % (gene_class - 2) + 1;
    P1 = timesPiont(k, G);
    num_t = rand() % eccPoint.len; //选择映射点
    Pt = eccPoint.p[num_t];
    //printf("Pt:(%d,%d)\n",Pt.point_x,Pt.point_y);
    P2 = timesPiont(k, PB);
    Pm = add_two_points(Pt, P2);
    printf("加密数据：\n");
    printf("kG=(%d,%d),Pt+kPB=(%d,%d),C={", P1.point_x, P1.point_y, Pm.point_x, Pm.point_y);
    for (i = 0; i < strlen(plain); i++){
        C[i] = m[i] * Pt.point_x + Pt.point_y;
        printf("{%d}", C[i]);
    }
    printf("}\n");
}
//解密 
void decrypt_ecc(){
    Point temp, temp1;
    int m, i;
    temp = timesPiont(nB, P1);
    temp.point_y = 0 - temp.point_y;
    temp1 = add_two_points(Pm, temp);//求解Pt 
    //	printf("(%d,%d)\n",temp.point_x,temp.point_y);
    //	printf("(%d,%d)\n",temp1.point_x,temp1.point_y);
    printf("\n解密结果:");
    for (i = 0; i < strlen(plain); i++){
        m = (C[i] - temp1.point_y) / temp1.point_x;
        printf("%c", alphabet[m]);//输出密文 
    }
    printf("\n");
}
int main(){
    get_generetor_class();
    encrypt_ecc();
    decrypt_ecc();
    return 0;
}