求RSA算法的源代码(c语言)

Python016

求RSA算法的源代码(c语言),第1张

这个是我帮个朋友写的,写的时候发现其实这个没那么复杂,不过,时间复杂度要高于那些成型了的,为人所熟知的rsa算法的其他语言实现.

#include

int

candp(int

a,int

b,int

c)

{

int

r=1

b=b+1

while(b!=1)

{

r=r*a

r=r%c

b--

}

printf("%d",r)

return

r

}

void

main()

{

int

p,q,e,d,m,n,t,c,r

char

s

{printf("input

the

p:\n")

scanf("%d\n",&p)

printf("input

the

q:\n")

scanf("%d%d\n",&p)

n=p*q

printf("so,the

n

is

%3d\n",n)

t=(p-1)*(q-1)

printf("so,the

t

is

%3d\n",t)

printf("please

intput

the

e:\n")

scanf("%d",&e)

if(e<1||e>t)

{printf("e

is

error,please

input

again")

scanf("%d",&e)}

d=1

while

(((e*d)%t)!=1)

d++

printf("then

caculate

out

that

the

d

is

%5d",d)

printf("if

you

want

to

konw

the

cipher

please

input

1\n

if

you

want

to

konw

the

plain

please

input

2\n")

scanf("%d",&r)

if(r==1)

{

printf("input

the

m

:"

)/*输入要加密的明文数字*/

scanf("%d\n",&m)

c=candp(m,e,n)

printf("so

,the

cipher

is

%4d",c)}

if(r==2)

{

printf("input

the

c

:"

)/*输入要解密的密文数字*/

scanf("%d\n",&c)

m=candp(c,d,n)

printf("so

,the

cipher

is

%4d\n",m)

printf("do

you

want

to

use

this

programe:yes

or

no")

scanf("%s",&s)

}while(s=='y')

}

}

程序修改如下:

(主要是你的循环写的不对,输入的字符应该-'0'才能与正常的数字对应)

#include<stdio.h>

#include<math.h>

int

candp(int

a,int

b,int

c)

{int

r=1

int

s

int

i=1

for(i=1i<=bi++)r=r*a

printf("%d\

",r)

s=r%c

printf("%d\

",s)

return

s}

void

main()

{

int

p,q,e,d,m,n,t,c,r

char

s

printf("please

input

the

p,q:")

scanf("%d%d",&p,&q)

n=p*q

t=(p-1)*(q-1)

printf("the

n

is

%12d\

",n)

printf("please

input

the

e:")

scanf("%d",&e)

while(e<1||e>n)

//此处修改为while循环

{

printf("e

is

error,please

input

again:")

scanf("%d",&e)

}

d=1

while(((e*d)%t)!=1)

d++

printf("then

caculate

out

that

the

d

is

%d\

",d)

printf("the

cipher

please

input

1\

")

printf("the

plain

please

input

2\

")

scanf("%c",&s)

while((s-'0')!=1&&(s-'0')!=2)

//消除后面的getchar()

此处增加while循环注意括号内的字符

{scanf("%c",&s)}

switch(s-'0')

{

case

1:printf("intput

the

m:")

scanf("%d",&m)

c=candp(m,e,n)

printf("the

plain

is

%d\

",c)break

case

2:printf("input

the

c:")

scanf("%d",&c)

m=candp(c,d,n)

printf("the

cipher

is

%8d\

",m)

break

}

}

#include <stdlib.h>

#include <stdio.h>

#include <string.h>

#include <math.h>

#include <time.h>

#define PRIME_MAX 200   // 生成素数范围

#define EXPONENT_MAX 200 // 生成指数e范围

#define Element_Max 127    // 加密单元的最大值,这里为一个char, 即1Byte

char str_read[100]="hello world !"  // 待加密的原文

int str_encrypt[100]                // 加密后的内容

char str_decrypt[100]              // 解密出来的内容

int str_read_len                    // str_read 的长度

int prime1, prime2                  // 随机生成的两个质数

int mod, eular                      // 模数和欧拉数

int pubKey, priKey                  // 公钥指数和私钥指数

// 生成随机素数,实际应用中,这两个质数越大,就越难破解。

int randPrime()

{

int prime, prime2, i

next:

prime = rand() % PRIME_MAX   // 随机产生数

if (prime <= 1) goto next      // 不是质数,生成下一个随机数

if (prime == 2 || prime == 3) return prime

prime2 = prime / 2              // prime>=4, prime2 的平方必定大于 prime , 因此只检查小于等于prime2的数

for (i = 2i <= prime2i++)   // 判断是否为素数

{

if (i * i >prime) return prime

if (prime % i == 0) goto next  // 不是质数,生成下一个随机数

}

}

// 欧几里德算法,判断a,b互质

int gcd(int a, int b)

{

int temp

while (b != 0) {

temp = b

b = a % b

a = temp

}

return a

}

//生成公钥指数,条件是 1<e <欧拉数,且与欧拉数互质。

int randExponent()

{

int e

while (1)

{

e = rand() % eularif (e <EXPONENT_MAX) break

}

while (1)

{

if (gcd(e, eular) == 1) return ee = (e + 1) % eularif (e == 0 || e >EXPONENT_MAX) e = 2

}

}

//生成私钥指数

int inverse()

{

int d, x

while (1)

{

d = rand() % eular

x = pubKey * d % eular

if (x == 1)

{

return d

}

}

}

//加密函数

void jiami()           

{

str_read_len = strlen(str_read)      //从参数表示的地址往后找,找到第一个'\0',即串尾。计算'\0'至首地址的“距离”,即隔了几个字符,从而得出长度。

printf("密文是:")

for (int i = 0i <str_read_leni++)

{

int C = 1int a = str_read[i], b = a % mod

for (int j = 0j <pubKeyj++) //实现加密

{

C = (C*b) % mod

}

str_encrypt[i] = C

printf("%d ", str_encrypt[i])

}

printf("\n")

}

//解密函数

void jiemi()         

{

int i=0  for (i = 0i <str_read_leni++)

{

int C = 1int a = str_encrypt[i], b=a%mod

for (int j = 0j <priKeyj++)

{

C = (C * b) % mod

}

str_decrypt[i] = C

}

str_decrypt[i] = '\0'printf("解密文是:%s \n", str_decrypt)

}

int main()

{

srand(time(NULL))

while (1)

{

prime1 = randPrime()prime2 = randPrime()printf("随机产生两个素数:prime1 = %d , prime2 = %d ", prime1, prime2)

mod = prime1 * prime2printf("模数:mod = prime1 * prime2 = %d \n", mod)if (mod >Element_Max) break// 模数要大于每个加密单元的值

}

eular = (prime1 - 1) * (prime2 - 1)  printf("欧拉数:eular=(prime1-1)*(prime2-1) = %d \n", eular)

pubKey = randExponent()printf("公钥指数:pubKey = %d\n", pubKey)

priKey = inverse()printf("私钥指数:priKey = %d\n私钥为 (%d, %d)\n", priKey, priKey, mod)

jiami()jiemi()

return 0

}