c语言编写hill密码

花了些时间写的，希望对你有帮助~

#include<stdio.h>

#define N 3 //可加密举洞的字符串长度

char plaintext[N]= //明文，输入时输入字符，参与运算时强制转换成整数

int ciphertext[N]=//密文，保存成整数，输出时强制转换成字符

int key[N][N]//密钥矩阵

void getPlainText() //获得明文字符串

{

printf("请输入明文:")

scanf("%s",plaintext)

printf("\数锋n")

}

void getKey()//输入密钥矩阵

{

int i,j

printf("请输入加密矩阵:\n")

for(i=0i<Ni++)

for(j=0j<Nj++)

scanf("%d",&key[i][j])

printf("\n")

}

void hill() //Hill加密算法

{

int i,j

for(i=0i<Ni++){ //矩阵的乘法

for(j=0j<Nj++){ //计算转换依据，0~25对应a~z

ciphertext[i]+=key[i][j]*(int)(plaintext[j]-'a')

ciphertext[i]%=26

}

}

printf("\n")

}

void printCipherText()//输出加密后的密文

{

int i

printf("加密后的密文是:"薯答晌)

for(i=0i<Ni++) //把参与计算后是整数强制转换成对应的字符

printf("%c",(char)(ciphertext[i]+'a'))

printf("\n")

}

void main()

{

getPlainText()//明文

getKey()//密钥

hill() //加密

printCipherText()//密文

}

这是C的腔昌，且基于的是ASCII码，大致改一改吧。

#include<stdio.h>

#include<math.h>

#include<stdlib.h>

#include<轿迟string.h>

struct hill

{

int i,j

int s[10][10]

}a[3],b

struct hill mul(struct hill,struct hill)

int det(struct hill,int,int)

struct hill inv(struct hill)

void code()

void uncode()

FILE *word,*num

void main()

{

int i=0,j=0,k=0

word=fopen("word.txt","r")

num=fopen("num.txt","r")

if(word==NULL||num==NULL)

{

printf("Error!\n")

exit(0)

}

fclose(word)

fclose(num)

while(!i)

{

fflush(stdin)

printf("输入密码\n")

for(i=0i<10i++)

{

for(j=0j<10j++)

{

a[k].s[i][j]=getchar()

if(a[k].s[i][j]=='\n') break

}

if(a[k].s[i][0]=='\n') break

}

for(a[k].j=0a[k].s[0][a[k].j]!='\n'a[k].j++)

for(a[k].i=0a[k].s[a[k].i][0]!='\n'a[k].i++) a[k].s[a[k].i][a[k].j]=0

for(i=0i<a[k].ii++) for(j=0j<a[k].jj++) a[k].s[i][j]=a[k].s[i][j]-'a'

i=0

if(a[k].i==a[k].j)

{

if(det(a[k],a[k].i,a[k].j)%2!=0&&det(a[k],a[k].i,a[k].j)%13!=0)

{

printf("是(0)否(1)继续输入?:")

scanf("%d"伍帆扒,&i)

k++

}

else printf("密码不合格!\n")

}

else printf("密码不合格!\n")

}

while(i)

{

printf("1:加密 2:解密\n")

scanf("%d",&i)

switch(i)

{

case 1: code()break

case 2: uncode()break

default: i=0

}

}

}

struct hill mul(struct hill a,struct hill b)

{

int i=0,j=0,k=0

struct hill ans

ans.i=a.i

ans.j=b.j

for(i=0i<ans.ii++)

{

for(j=0j<ans.jj++)

{

ans.s[i][j]=0

for(k=0k<a.jk++)

ans.s[i][j]=ans.s[i][j]+a.s[i][k]*b.s[k][j]

}

}

for(i=0i<ans.ii++)

{

for(j=0j<ans.jj++) ans.s[i][j]=(ans.s[i][j]%26+26)%26+'a'

ans.s[i][j]='\0'

}

return(ans)

}

struct hill inv(struct hill a)

{

char c

int i=0,j=0,k=0

struct hill ans

ans.i=a.i

ans.j=a.j

switch(det(a,a.i,a.j)%26)

{

case 1: k=1break

case 3: k=9break

case 5: k=21break

case 7: k=15break

case 9: k=3break

case 11: k=19break

case 15: k=7break

case 17: k=23break

case 19: k=11break

case 21: k=5break

case 23: k=17break

case 25: k=25break

}

for(i=0i<ans.ii++)

{

for(j=0j<ans.jj++)

ans.s[i][j]=(((int)pow(-1,i+j)*det(a,i,j)*k)%26+26)%26

}

for(i=0i<ans.ii++) ans.s[i][ans.j]='\0'

for(i=0i<ans.ii++)

{

for(j=0j<ij++)

{

c=ans.s[i][j]

ans.s[i][j]=ans.s[j][i]

ans.s[j][i]=c

}

}

return(ans)

}

int det(struct hill a,int m,int n)

{

int i=0,j=0,x=0

if(m!=a.i||n!=a.j)

{

for(i=0i<a.ii++) for(j=nj<a.jj++) a.s[i][j]=a.s[i][j+1]

for(i=mi<a.ii++) for(j=0j<a.jj++) a.s[i][j]=a.s[i+1][j]

a.i--

a.j--

}

if(a.i==2&&a.j==2) return(a.s[0][0]*a.s[1][1]-a.s[0][1]*a.s[1][0])

for(i=0i<a.ii++) x+=a.s[0][i]*det(a,0,i)*(int)pow(-1,i)

return(x)

}

void code()

{

int i=0,j=0,space[10],k=0,n=0

struct hill ans

char c=0

word=fopen("word.txt","r")

c=fgetc(word)

while(c!=EOF)

{

if(c>='a'&&c<='z') n++

c=fgetc(word)

}

fclose(word)

word=fopen("word.txt","r")

num=fopen("num.txt","w")

while(n)

{

printf("剩余%d: ",n)

scanf("%d",&j)

n=n-j

for(k=0a[k].i!=jj++)

b.i=a[k].i

b.j=1

for(j=0j<10j++) space[j]=-1

j=0

for(i=0i<b.ii++)

{

b.s[i][0]=fgetc(word)

if(b.s[i][0]<'a'||b.s[i][0]>'z')

{

space[j]=i

j++

b.s[i][0]=fgetc(word)

}

}

for(i=0i<b.ii++)

{

b.s[i][0]-='a'

b.s[i][1]='\0'

}

ans=mul(a[k],b)

for(i=0i<ans.ii++)

{

for(j=0j<10j++) if(i==space[j]) break

if(j!=10) fprintf(num," %c",ans.s[i][0])

else fprintf(num,"%c",ans.s[i][0])

}

}

fclose(word)

fclose(num)

}

void uncode()

{

int k=0,n=0

int i=0,j=0,space[10]

struct hill ans

char c=0

word=fopen("num.txt","r")

c=fgetc(word)

while(c!=EOF)

{

if(c>='a'&&c<='z') n++

c=fgetc(word)

}

fclose(word)

word=fopen("word.txt","w")

num=fopen("num.txt","r")

b.i=a[k].i

b.j=1

while(n)

{

printf("剩余%d: ",n)

scanf("%d",&j)

n=n-j

for(k=0a[k].i!=jj++)

b.i=a[k].i

b.j=1

for(j=0j<10j++) space[j]=-1

j=0

for(i=0i<b.ii++)

{

b.s[i][0]=fgetc(num)

if(b.s[i][0]<'a'||b.s[i][0]>'z')

{

space[j]=i

j++

b.s[i][0]=fgetc(num)

}

}

for(i=0i<b.ii++)

{

b.s[i][0]-='a'

b.s[i][1]='\0'

}

ans=mul(inv(a[k]),b)

for(i=0i<ans.ii++)

{

for(j=0j<10j++) if(i==space[j]) break

if(j!=10) fprintf(word," %c",ans.s[i][0])

else fprintf(word,"%c",ans.s[i][0])

}

}

fclose(word)

fclose(num)

}

// 希尔算法的加密与解密\x0d\x0a#include \x0d\x0a#include \x0d\x0a#include \x0d\x0a#include \x0d\x0a#include // nDime为全部变量，可逆矩阵的维数\x0d\x0aint nDime\x0d\x0aint index = 0// MAXN为明文的最大长度\x0d\x0aconst int MAXN = 256// 矩阵相乘，a是一个列为早掘扒1的矩阵\x0d\x0avoid MultiplyMatrix(int a[], int b[][10], int *text)\x0d\x0a{\x0d\x0a int i, j, tfor (i = 0i = size)\x0d\x0asum %= size\x0d\x0a temp1[diff] *= m[row][column]\x0d\x0a temp2[sum] *= m[row][column]\x0d\x0a }\x0d\x0a }\x0d\x0a t = 0\x0d\x0a for (row = 0row 回答于 2022-11-16

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