求RSA加解密系统的C++代码

要求素数P和Q为五位十进制p、q、e最好能自动生成。。

2025-02-12 21:46:48

推荐回答（1个）

回答1：

//下面程序由520huiqin编写，已在VC++ 6.0下编译通过

#include
#include
#include

typedef int Elemtype;
Elemtype p,q,e;
Elemtype fn;
Elemtype m,c;
int flag = 0;
typedef void (*Msghandler) (void);
struct MsgMap {
char ch;
Msghandler handler;
};
/* 公钥 */
struct PU {
Elemtype e;
Elemtype n;
} pu;
/* 私钥 */
struct PR {
Elemtype d;
Elemtype n;
} pr;
/* 判定一个数是否为素数 */
bool test_prime(Elemtype m) {
if (m <= 1) {
return false;
}
else if (m == 2) {
return true;
}
else {
for(int i=2; i<=sqrt(m); i++) {
if((m % i) == 0) {
return false;
break;
}
}
return true;
}
}
/* 将十进制数据转化为二进制数组 */
void switch_to_bit(Elemtype b, Elemtype bin[32]) {
int n = 0;
while( b > 0) {
bin[n] = b % 2;
n++;
b /= 2;
}
}
/* 候选菜单，主界面 */
void Init() {
cout<<"*********************************************"<cout<<"*** Welcome to use RSA encoder ***"<cout<<"*** a.about ***"<cout<<"*** e.encrypt ***"<cout<<"*** d.decrypt ***"<cout<<"*** s.setkey ***"<cout<<"*** q.quit ***"<cout<<"**********************************by*Terry***"<cout<<"press a key:"<}
/* 将两个数排序，大的在前面*/
void order(Elemtype &in1, Elemtype &in2) {
Elemtype a = ( in1 > in2 ? in1 : in2);
Elemtype b = ( in1 < in2 ? in1 : in2);
in1 = a;
in2 = b;
}
/* 求最大公约数 */
Elemtype gcd(Elemtype a, Elemtype b) {
order(a,b);
int r;
if(b == 0) {
return a;
}
else {
while(true) {
r = a % b;
a = b;
b = r;
if (b == 0) {
return a;
break;
}
}
}

}
/* 用扩展的欧几里得算法求乘法逆元 */
Elemtype extend_euclid(Elemtype m, Elemtype bin) {
order(m,bin);
Elemtype a[3],b[3],t[3];
a[0] = 1, a[1] = 0, a[2] = m;
b[0] = 0, b[1] = 1, b[2] = bin;
if (b[2] == 0) {
return a[2] = gcd(m, bin);
}
if (b[2] ==1) {
return b[2] = gcd(m, bin);
}
while(true) {
if (b[2] ==1) {
return b[1];
break;
}
int q = a[2] / b[2];
for(int i=0; i<3; i++) {
t[i] = a[i] - q * b[i];
a[i] = b[i];
b[i] = t[i];
}
}
}
/* 快速模幂算法 */
Elemtype modular_multiplication(Elemtype a, Elemtype b, Elemtype n) {
Elemtype f = 1;
Elemtype bin[32];
switch_to_bit(b,bin);
for(int i=31; i>=0; i--) {
f = (f * f) % n;
if(bin[i] == 1) {
f = (f * a) % n;
}
}
return f;
}
/* 产生密钥 */
void produce_key() {
cout<<"input two primes p and q:";
cin>>p>>q;
while (!(test_prime(p)&&test_prime(q))){
cout<<"wrong input,please make sure two number are both primes!"< cout<<"input two primes p and q:";
cin>>p>>q;
};
pr.n = p * q;
pu.n = p * q;
fn = (p - 1) * (q - 1);
cout<<"fn = "<cout<<"input e :";
cin>>e;
while((gcd(fn,e)!=1)) {
cout<<"e is error,try again!";
cout<<"input e :";
cin>>e;
}
pr.d = (extend_euclid(fn,e) + fn) % fn;
pu.e = e;
flag = 1;
cout<<"PR.d: "<cout<<"PU.e: "<}
/* 加密 */
void encrypt() {
if(flag == 0) {
cout<<"setkey first:"< produce_key();
}
cout<<"input m:";
cin>>m;
c = modular_multiplication(m,pu.e,pu.n);
cout<<"c is:"<}
/* 解密 */
void decrypt() {
if(flag == 0) {
cout<<"setkey first:"< produce_key();
}
cout<<"input c:";
cin>>c;
m = modular_multiplication(c,pr.d,pr.n);
cout<<"m is:"<}
/* 版权信息 */
void about() {
cout<<"*********************************************"<cout<<"*** by Terry ***"<cout<<"*** copyright 2010,All rights reserved by ***"<cout<<"*** Terry,technology supported by weizuo !***"<cout<<"*** If you have any question, please mail ***"<cout<<"*** to 18679376@qq.com ! ***"<cout<<"*** Computer of science and engineering ***"<cout<<"*** XiDian University 2010-4-29 ***"<cout<<"*********************************************"<cout<Init();
}
/* 消息映射 */
MsgMap Messagemap[] = {
{'a',about},
{'s',produce_key},
{'d',decrypt},
{'e',encrypt},
{'q',NULL}
};
/* 主函数，提供循环 */
void main() {
Init();
char d;
while((d = getchar())!='q') {
int i = 0;
while(Messagemap[i].ch) {
if(Messagemap[i].ch == d) {
Messagemap[i].handler();
break;
}
i++;
}
}
}

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