C语言加密算法实现（乘数密码、维吉尼亚密码、DES密码）

C语言加密算法实现

任务一、乘数密码算法实现

源码：

#include
#include
#include
#define MOD_NUM 26
#define ENCRY_DECRY_SIZE 500
int euclidean(int d, int f);
char char_encrypting(char a, int secret);
char char_decrypting(char ciphertext, int secret);
char* string_encrypting(char* plaintext, int secret);
char* string_decrypting(char ciphertext[], int secret);

int main()
{
	int secret = 3; //密钥
	char* plaintext; //明文
	char* ciphertext; //密文
	printf("请输入要加密的内容:");
	scanf("%s", plaintext);
	ciphertext = string_encrypting(plaintext, secret);
	printf("加密后的密文是:%s\n", ciphertext);	
	printf("请输入要解密的内容:");
	scanf("%s", ciphertext);
	plaintext = string_decrypting(ciphertext, secret);
	printf("解密后的密文是:%s\n", plaintext);
}
//字符串加密
char* string_encrypting(char plaintext[], int secret)
{
	char* ciphertext;
	ciphertext = (char*) malloc (sizeof(char) * ENCRY_DECRY_SIZE);
	for(int i = 0; i < strlen(plaintext); i++) {
		ciphertext[i] = char_encrypting(plaintext[i], secret);
		/*if(ciphertext[i] == 'a' - 1)
			ciphertext[i] = 'a';*/
	}
	ciphertext[strlen(plaintext)] = '\0';
	return ciphertext;
}
//字符串解密
char* string_decrypting(char ciphertext[], int secret)
{
	char* plaintext;
	plaintext = (char*) malloc (sizeof(char) * ENCRY_DECRY_SIZE);
	for(int i = 0; i < strlen(ciphertext); i++) {
		plaintext[i] = char_decrypting(ciphertext[i], secret);
	}
	plaintext[strlen(ciphertext)] = '\0';
	return plaintext;
}
//字符加密
char char_encrypting(char a, int secret)
{
	return (a-'a') * secret % MOD_NUM + 'a';
}
//字符解密
char char_decrypting(char ciphertext, int secret)
{
	return (ciphertext-'a') * euclidean(secret, MOD_NUM) % MOD_NUM + 'a';
}
//欧几里得算法，用于计算乘法逆元
int euclidean(int d, int f)
{
	//1
	int x[3], y[3];
	x[0] = 1; x[1] = 0; x[2] = f;
	y[0] = 0; y[1] = 1; y[2] = d;
	//2
flag:	if(y[2] == 0)
		return -1; //null
	//3
	if(y[2] == 1)
		return y[1];
	//4
	int q = x[2] / y[2];
	int T[3] = { x[0]-q*y[0], x[1]-q*y[1], x[2]-q*y[2]};
	x[0] = y[0]; x[1] = y[1]; x[2] = y[2];
	y[0] = T[0]; y[1] = T[1]; y[2] = T[2];
	goto flag;
}

结果截图：

任务二、维吉尼亚密码算法

源代码：

#include
#include
#define MOD_SIZE 26
/*
加密函数：
Ci=Pi+Ki(mod 26)
解密函数：
Pi=Ci-Ki(mod 26)
*/
//加密函数
char encrypting(char plaintext_c, char secret)
{
	char result = ((secret-'a') % MOD_SIZE + plaintext_c - 'a') % MOD_SIZE + 'A';
	return result;
}
int main()
{
	int count = 0;
	char sec_c;
	const char* secret = "better\0";//密钥
	const char* plaintext = "we will have a party on friday\0";//需要加密的明文
	for(int i = 0;i < strlen(plaintext); i++) {
		if(plaintext[i] == ' ')
			printf(" ");
		else
		{
			sec_c = secret[count++];
			printf("%c", encrypting(plaintext[i], sec_c));
            //循环使用密钥
			count = count % strlen(secret);
		}
	}
}

结果截图

任务三、部分DES加密算法实现

参考博客：https://blog.csdn.net/dyw_666666/article/details/85719984

源代码：

#include
#include
//置换选择PC-1
int pc_1[60]= {0,57,49,41,33,25,17,9,
               1,58,50,42,34,26,18,
               10,2,59,51,43,35,27,
               19,11,3,60,52,44,36,
               63,55,47,39,31,23,15,
               7,62,54,46,38,30,22,
               14,6,61,53,45,37,29,
               21,13,5,28,20,12,4};

//置换选择PC-2
int pc_2[60]= {0,14,17,11,24,1,5,
               3,28,15,6,21,10,
               23,19,12,4,26,8,
               16,7,27,20,13,2,
               41,52,31,37,47,55,
               30,40,51,45,33,48,
               44,49,39,56,34,53,
               46,42,50,36,29,32};
//IP初始置换表
int pc_ip[80]= {0,58,50,42,34,26,18,10,2,
                60,52,44,36,28,20,12,4,
                62,54,46,38,30,22,14,6,
                64,56,48,40,32,24,16,8,
                57,49,41,33,25,17,9,1,
                59,51,43,35,27,19,11,3,
                61,53,45,37,29,21,13,5,
                63,55,47,39,31,23,15,7};

//逆初始置换表
int pc_ip_1[80]= {0,40,8,48,16,56,24,64,32,
                  39,7,47,15,55,23,63,31,
                  38,6,46,14,54,22,62,30,
                  37,5,45,13,53,21,61,29,
                  36,4,44,12,52,20,60,28,
                  35,3,43,11,51,19,59,27,
                  34,2,42,10,50,18,58,26,
                  33,1,41,9,49,17,57,25};
//S盒
int s_box[8][4][16]={
	14,4,13,1,2,15,11,8,3,10,6,12,5,9,0,7,
	0,15,7,4,14,2,13,1,10,6,12,11,9,5,3,8,
	4,1,14,8,13,6,2,11,15,12,9,7,3,10,5,0,
	15,12,8,2,4,9,1,7,5,11,3,14,10,0,6,13,
 
	15,1,8,14,6,11,3,4,9,7,2,13,12,0,5,10,
	3,13,4,7,15,2,8,14,12,0,1,10,6,9,11,5,
	0,14,7,11,10,4,13,1,5,8,12,6,9,3,2,15,
	13,8,10,1,3,15,4,2,11,6,7,12,10,5,14,9,
 
	10,0,9,14,6,3,15,5,1,13,12,7,11,4,2,8,
	13,7,0,9,3,4,6,10,2,8,5,14,12,11,15,1,
	13,6,4,9,8,15,3,0,11,1,2,12,5,10,14,7,
	1,10,13,0,6,9,8,7,4,15,14,3,11,5,2,12,
 
	7,13,14,3,0,6,9,10,1,2,8,5,11,12,4,15,
	13,8,11,5,6,15,0,3,4,7,2,12,1,10,14,9,
	10,6,9,0,12,11,7,13,15,1,3,14,5,2,8,4,
	3,15,0,6,10,1,13,8,9,4,5,11,12,7,2,14,
 
	2,12,4,1,7,10,11,6,8,5,3,15,13,0,14,9,
	14,11,2,12,4,7,13,1,5,0,15,10,3,9,8,6,
	4,2,1,11,10,13,7,8,15,9,12,5,6,3,0,14,
	11,8,12,7,1,14,2,13,6,15,0,9,10,4,5,3,
 
	12,1,10,15,9,2,6,8,0,13,3,4,14,7,5,11,
	10,15,4,2,7,12,9,5,6,1,13,14,0,11,3,8,
	9,14,15,5,2,8,12,3,7,0,4,10,1,13,11,6,
	4,3,2,12,9,5,15,10,11,14,1,7,6,0,8,13,
 
	4,11,2,14,15,0,8,13,3,12,9,7,5,10,6,1,
	13,0,11,7,4,9,1,10,14,3,5,12,2,15,8,6,
	1,4,11,13,12,3,7,14,10,15,6,8,0,5,9,2,
	6,11,13,8,1,4,10,7,9,5,0,15,14,2,3,12,
 
	13,2,8,4,6,15,11,1,10,9,3,14,5,0,12,7,
	1,15,13,8,10,3,7,4,12,5,6,11,0,14,9,2,
	7,11,4,1,9,12,14,2,0,6,10,13,15,3,5,8,
	2,1,14,7,4,10,8,13,15,12,9,0,3,5,6,11};
//E盒扩展变换
int pc_e[80]= {0,32,1,2,3,4,5,
               4,5,6,7,8,9,
               8,9,10,11,12,13,
               12,13,14,15,16,17,
               16,17,18,19,20,21,
               20,21,22,23,24,25,
               24,25,26,27,28,29,
               28,29,30,31,32,1};
 
//P盒置换表
int pc_p[80]= {0,16,7,20,21,
               29,12,28,17,
               1,15,23,26,
               5,18,31,10,
               2,8,24,14,
               32,27,3,9,
               19,13,30,6,
               22,11,4,25};
struct node
{
    int c[80];
    int d[80];
    int cd[80];
    int k_n[80];
    int l[80];
    int r[80];
    node()
    {
        memset(c,0,sizeof(c));
        memset(d,0,sizeof(d));
        memset(k_n,0,sizeof(k_n));
        memset(l,0,sizeof(l));
        memset(r,0,sizeof(r));
    }
};
char plaintext[16] = {'3','1','3','2','3','3','3','4','3','5','3','6','3','7','3','8'};
char k[18] = {'3','1','3','2','3','3','3','4','3','5','3','6','3','7','3','8'};
///F函数实现E盒扩展
int f(int r[80],int kn[80])
{
    int x=0;
    int e[80]= {0};
    int h=0,l=0,idx=0;
    //E盒扩展得到e(Ri);
    for(int i=1; i<=48; i++)
        e[i]=r[pc_e[i]];
    //异或运算得到k^(Ri);
    for(int i=1; i<=48; i++)
        r[i]=e[i]^kn[i];
    //S盒压缩将48位分为8组，每组6位，得到S(k^(Ri));
    for(int i=1; i<=48; i+=6)
    {
        h=r[i]*2+r[i+5]*1;
        l=r[i+1]*8+r[i+2]*4+r[i+3]*2+r[i+4]*1;
        e[++idx]=(s_box[x][h][l]>>3)&1;
        e[++idx]=(s_box[x][h][l]>>2)&1;
        e[++idx]=(s_box[x][h][l]>>1)&1;
        e[++idx]=s_box[x][h][l]&1;
        x++;
    }
    //P盒置换
    for(int i=1; i<=32; i++)
        r[i]=e[pc_p[i]];
}
int jiami(char m[])
{
	int two[66] = {0};
	int two_k[66] = {0};
	int two_kup[66] = {0};
	int idx = 0;
	int num = 0;
	//明文十六进制转为二进制
	for(int i = 0; i < 16; i++)
	{
		if(m[i] >= '0' && m[i] <= '9')
			num = m[i] - '0';
		else
			num = m[i] - 'A' + 10;
		two[++idx] = (num>>3)&1;
		two[++idx] = (num>>2)&1;
		two[++idx] = (num>>1)&1;
		two[++idx] = num&1;
	}
	idx = 0;
	//密钥转2进制
	for(int i = 0;i < 16; i++)
	{
		if(k[i]>='0'&&k[i]<='9')
			num=k[i]-'0';
		else num=k[i]-'A'+10;
		two_k[++idx]=(num>>3)&1;
		two_k[++idx]=(num>>2)&1;
		two_k[++idx]=(num>>1)&1;
		two_k[++idx]=num&1;
	}
	//迭代求出１６个密钥
	//用pc_1得到k+(将密钥用pc_1置换)
	for(int i=1; i<=56; i++)
		two_kup[i]=two_k[pc_1[i]];
	//左右分组
	node c_and_d[20];
	for(int i=1; i<=28; i++) //得到c[0];
	c_and_d[0].c[i]=two_kup[i];
	for(int i=1,j=29; j<=56; i++,j++) //得到d[0];
	c_and_d[0].d[i]=two_kup[j];
	///16轮生成每轮子密钥
	for(int i=1; i<=16; i++) //得到c[1]-c[15],d[1]-d[15],CnDn;
	{
		//如果为第1、2、9、16轮，则d循环左移1位
		if(i==1||i==2||i==9||i==16)
		{
			for(int j=1; j<=27; j++)
				c_and_d[i].c[j]=c_and_d[i-1].c[j+1];
			c_and_d[i].c[28]=c_and_d[i-1].c[1];
			for(int j=1; j<=27; j++)
				c_and_d[i].d[j]=c_and_d[i-1].d[j+1];
			c_and_d[i].d[28]=c_and_d[i-1].d[1];
		}
		//如果是其他轮次，则d循环左移2位
		else
		{
			for(int j=1; j<=26; j++)
				c_and_d[i].c[j]=c_and_d[i-1].c[j+2];
			c_and_d[i].c[27]=c_and_d[i-1].c[1];
			c_and_d[i].c[28]=c_and_d[i-1].c[2];
			for(int j=1; j<=26; j++)
				c_and_d[i].d[j]=c_and_d[i-1].d[j+2];
			c_and_d[i].d[27]=c_and_d[i-1].d[1];
			c_and_d[i].d[28]=c_and_d[i-1].d[2];
		}
		for(int j=1; j<=28; j++)
			c_and_d[i].cd[j]=c_and_d[i].c[j];
		for(int j=29,t=1; t<=28; j++,t++)
			c_and_d[i].cd[j]=c_and_d[i].d[t];
	}
	//统一将１６个左移后的密钥进行PC2置换
	printf("得到的１６轮密钥分别是:\n");
	node k_16[20];
	for(int i=1; i<=16; i++) //得到k1-kn;
	{
		for(int j=1; j<=48; j++)
		{ 
			k_16[i].k_n[j]=c_and_d[i].cd[pc_2[j]];
			printf("%d", k_16[i].k_n[j]);
		}
		printf("\n");
	}
	/*
	   得到１６个密钥后
	   DES算法明文加密
	   功能：实现DES算法对明文的16轮加密
	 */
	int ip[80]= {0};
	//１、初始置换ip
	for(int i=1; i<=64; i++)
		ip[i]=two[pc_ip[i]];
	
	//２、进行１６轮迭代
	//初始化得到 l[0],r[0]
	node l_r[20];
	for(int i=1; i<=32; i++) //得到l;
	l_r[0].l[i]=ip[i];
	for(int i=1,j=33; j<=64; i++,j++) //得到r;
	l_r[0].r[i]=ip[j];
	//开始１６轮迭代,计算L1-L16，R1-R16;
        for(int i=1; i<=16; i++)     
	{
            for(int j=1; j<=32; j++)
	    //每轮的左32位等于上一轮的右32位
                l_r[i].l[j]=l_r[i-1].r[j];
            ///F函数包含E盒扩展、异或、S盒压缩、P盒置换
            f(l_r[i-1].r,k_16[i].k_n);
            ///左右合在一起，两者进行最终按位异或得到r[i]
            for(int j=1; j<=32; j++)
                l_r[i].r[j]=l_r[i-1].l[j]^l_r[i-1].r[j];
        }

    int R16L16[80]= {0}; //得到R16L16;
    //F函数过后左右合在一起
    for(int i=1; i<=32; i++)
        R16L16[i]=l_r[16].r[i];
    for(int i=33,j=1; j<=32; j++,i++)
        R16L16[i]=l_r[16].l[j];
 
    int ans[80]= {0}; //得到最终变换;
printf("最终得到的密文:\n");
    //进行ip逆置换
    for(int i=1; i<=64; i++)
    { 
 	  ans[i]=R16L16[pc_ip_1[i]];
	printf("%d", ans[i]);
	}
}
int main()
{
	jiami(plaintext);
}

结果截图：