❶ 如何用C語言實現RSA演算法
上學期交的作業，已通過老師在運行時間上的測試
#include <stdio.h>
#include <stdlib.h>
unsigned long prime1,prime2,ee;
unsigned long *kzojld(unsigned long p,unsigned long q) //擴展歐幾里得演算法求模逆
{
unsigned long i=0,a=1,b=0,c=0,d=1,temp,mid,ni[2];
mid=p;
while(mid!=1)
{
while(p>q)
{p=p-q; mid=p;i++;}
a=c*(-1)*i+a;b=d*(-1)*i+b;
temp=a;a=c;c=temp;
temp=b;b=d;d=temp;
temp=p;p=q;q=temp;
i=0;
}
ni[0]=c;ni[1]=d;
return(ni);
}
unsigned long momi(unsigned long a,unsigned long b,unsigned long p) //模冪演算法
{
unsigned long c;
c=1;
if(a>p) a=a%p;
if(b>p) b=b%(p-1);
while(b!=0)
{
while(b%2==0)
{
b=b/2;
a=(a*a)%p;
}
b=b-1;
c=(a*c)%p;
}
return(c);
}
void RSAjiami() //RSA加密函數
{
unsigned long c1,c2;
unsigned long m,n,c;
n=prime1*prime2;
system("cls");
printf("Please input the message：\n");
scanf("%lu",&m);getchar();
c=momi(m,ee,n);
printf("The cipher is：%lu",c);
return;
}
void RSAjiemi() //RSA解密函數
{
unsigned long m1,m2,e,d,*ni;
unsigned long c,n,m,o;
o=(prime1-1)*(prime2-1);
n=prime1*prime2;
system("cls");
printf("Please input the cipher：\n");
scanf("%lu",&c);getchar();
ni=kzojld(ee,o);
d=ni[0];
m=momi(c,d,n);
printf("The original message is：%lu",m);
return;
}
void main()
{ unsigned long m;
char cho;
printf("Please input the two prime you want to use:\n");
printf("P=");scanf("%lu",&prime1);getchar();
printf("Q=");scanf("%lu",&prime2);getchar();
printf("E=");scanf("%lu",&ee);getchar();
if(prime1<prime2)
{m=prime1;prime1=prime2;prime2=m;}
while(1)
{
system("cls");
printf("\t*******RSA密碼系統*******\n");
printf("Please select what do you want to do:\n");
printf("1.Encrpt.\n");
printf("2.Decrpt.\n");
printf("3.Exit.\n");
printf("Your choice:");
scanf("%c",&cho);getchar();
switch(cho)
{ case '1':RSAjiami();break;
case '2':RSAjiemi();break;
case '3':exit(0);
default:printf("Error input.\n");break;
}
getchar();
}
}
❷ 求正確的RSA加密解密演算法C語言的，多謝。
//rsa.h
#include<stdio.h>
#defineMAX_NUM63001
#defineMAX_PRIME251
//!返回代碼
#defineOK100
#defineERROR_NOEACHPRIME101
#defineERROR_NOPUBLICKEY102
#defineERROR_GENERROR103
unsignedintMakePrivatedKeyd(unsignedintuiP,unsignedintuiQ);
unsignedintGetPrivateKeyd(unsignedintiWhich);
unsignedintMakePairkey(unsignedintuiP,unsignedintuiQ,unsignedintuiD);
unsignedintGetPairKey(unsignedint&d,unsignedint&e);
voidrsa_encrypt(intn,inte,char*mw,intiLength,int*&cw);
voidrsa_decrypt(intn,intd,int*&cw,intcLength,char*mw);
voidoutputkey();
//rsa.c
#include"rsa.h"
//!保存私鑰d集合
structpKeyset
{
unsignedintset[MAX_NUM];
unsignedintsize;
}pset;
//!保存公、私鑰對
structpPairkey
{
unsignedintd;
unsignedinte;
unsignedintn;
}pairkey;
//名稱：isPrime
//功能：判斷兩個數是否互質
//參數：m:數a;n:數b
//返回：m、n互質返回true;否則返回false
boolisPrime(unsignedintm,unsignedintn)
{
unsignedinti=0;
boolFlag=true;
if(m<2||n<2)
returnfalse;
unsignedinttem=(m>n)?n:m;
for(i=2;i<=tem&&Flag;i++)
{
boolmFlag=true;
boolnFlag=true;
if(m%i==0)
mFlag=false;
if(n%i==0)
nFlag=false;
if(!mFlag&&!nFlag)
Flag=false;
}
if(Flag)
returntrue;
else
returnfalse;
}
//名稱：MakePrivatedKeyd
//功能：由素數Q、Q生成私鑰d
//參數：uiP:素數P;uiQ:素數Q
//返回：私鑰d
unsignedintMakePrivatedKeyd(unsignedintuiP,unsignedintuiQ)
{
unsignedinti=0;
//!得到所有與z互質的數(私鑰d的集合)
unsignedintz=(uiP-1)*(uiQ-1);
pset.size=0;
for(i=0;i<z;i++)
{
if(isPrime(i,z))
{
pset.set[pset.size++]=i;
}
}
returnpset.size;
}
//名稱：MakePairKey
//功能：生成RSA公、私鑰對
//參數：uiP:素數P;uiQ:素數Q;uiD:私鑰d
//返回：錯誤代碼
unsignedintMakePairkey(unsignedintuiP,unsignedintuiQ,unsignedintuiD)
{
boolbFlag=true;
unsignedinti=0,e;
unsignedintz=(uiP-1)*(uiQ-1);
unsignedintd=pset.set[uiD];
//d=uiD;
if(!isPrime(z,d))
returnERROR_NOEACHPRIME;
for(i=2;i<z;i++)
{
if((i*d)%z==1)
{
e=i;
bFlag=false;
}
}
if(bFlag)
returnERROR_NOPUBLICKEY;
if((d*e)%z!=1)
ERROR_GENERROR;
pairkey.d=d;
pairkey.e=e;
pairkey.n=uiP*uiQ;
returnOK;
}
//名稱：GetPairKey
//功能：對外提供介面，獲得公、私鑰對
//參數：uiP:素數P;uiQ:素數Q;uiD:私鑰d
//返回：
unsignedintGetPairKey(unsignedint&d,unsignedint&e)
{
d=pairkey.d;
e=pairkey.e;
returnpairkey.n;
}
//名稱：GetPrivateKeyd
//功能：對外提供介面，由用戶選擇ID得以私鑰d
//參數：iWhich:用戶選擇私鑰d的ID
//返回：私鑰d值
unsignedintGetPrivateKeyd(unsignedintiWhich)
{
if(pset.size>=iWhich)
returnpset.set[iWhich];
else
return0;
}
//名稱：rsa_encrypt
//功能：RSA加密運算
//參數：n:公鑰n;e:公鑰e;mw:加密明文;iLength:明文長度;cw:密文輸出
//返回：無
voidrsa_encrypt(intn,inte,char*mw,intmLength,int*&cw)
{
inti=0,j=0;
__int64temInt=0;
for(i=0;i<mLength;i++)
{
temInt=mw[i];
if(e!=0)
{
for(j=1;j<e;j++)
{
temInt=(temInt*mw[i])%n;
}
}
else
{
temInt=1;
}
cw[i]=(int)temInt;
}
}
//名稱：rsa_decrypt
//功能：RSA解密運算
//參數：n:私鑰n;d:私鑰d;cw:密文;cLength:密文長度;mw:明文輸出
//返回：無
voidrsa_decrypt(intn,intd,int*&cw,intcLength,char*mw)
{
inti=0,j=-1;
__int64temInt=0;
for(i=0;i<cLength/4;++i)
{
mw[i]=0;
temInt=cw[i];
if(d!=0)
{
for(j=1;j<d;j++)
{
temInt=(__int64)(temInt*cw[i])%n;
}
}
else
{
temInt=1;
}
mw[i]=(char)temInt;
}
}
voidoutputkey()
{
printf("PublicKey(e,n):(%d,%d) ",pairkey.e,pairkey.n);
printf("PrivateKey(d,n):(%d,%d) ",pairkey.d,pairkey.n);
}
//main.c
//工程：RSA
//功能：RSA加、解密文件
//作者：jlcss|ExpNIS
#include<stdio.h>
#include<afxwin.h>
#include<math.h>
#include"rsa.h"
#defineDECRYPT_FILE"RSA加密密文.txt"
#defineENCRYPT_FILE"RSA解密明文.txt"
//!約束文件最大2M
#defineMAX_FILE1024*1024*2
//名稱：usage
//功能：幫助信息
//參數：應用程序名稱
//返回：提示信息
voidUsage(constchar*appname)
{
printf(" usage:rsa-k素數P素數Q ");
printf(" usage:rsa-e明文文件公鑰e公鑰n ");
printf(" usage:rsa-d密文文件私鑰d私鑰n ");
}
//名稱：IsNumber
//功能：判斷數字字元數組
//參數：strNumber:字元數組
//返回：數字字組數組返回true，否則返回false;
boolIsNumber(constchar*strNumber)
{
unsignedinti;
if(!strNumber)
returnfalse;
for(i=0;i<strlen(strNumber);i++)
{
if(strNumber[i]<'0'||strNumber[i]>'9')
returnfalse;
}
returntrue;
}
//名稱：IsPrimeNumber
//功能：判斷素數
//參數：num:輸入整數
//返回：素數返回true，否則返回false;
boolIsPrimeNumber(unsignedintnum)
{
unsignedinti;
if(num<=1)
returnfalse;
unsignedintsqr=(unsignedint)sqrt((double)num);
for(i=2;i<=sqr;i++)
{
if(num%i==0)
returnfalse;
}
returntrue;
}
//名稱：FileIn
//功能：讀取磁碟文件到內存
//參數：strFile:文件名稱；inBuff:指向文件內容緩沖區
//返回：實際讀取內容大小(位元組)
intFileIn(constchar*strFile,unsignedchar*&inBuff)
{
intiFileLen=0,iBuffLen=0;
//!打開密文文件
CFilefile(strFile,CFile::modeRead);
iFileLen=(int)file.GetLength();
if(iFileLen>MAX_FILE)
{
printf("文件長度不能大於%dM,! ",MAX_FILE/(1024*1024));
gotoout;
}
iBuffLen=iFileLen;
inBuff=newunsignedchar[iBuffLen];
if(!inBuff)
gotoout;
ZeroMemory(inBuff,iBuffLen);
file.Read(inBuff,iFileLen);
file.Close();
out:
returniBuffLen;
}
//名稱：FileOut
//功能：加/解密結果輸出到當前目錄磁碟文件中
//參數：strOut指向輸出字元緩沖區，輸出大小len，strFile為輸出文件
//返回：無
voidFileOut(constvoid*strOut,intlen,constchar*strFile)
{
//!輸出到文件
CFileoutfile(strFile,CFile::modeCreate|CFile::modeWrite);
outfile.Write(strOut,len);
outfile.Close();
}
//名稱：CheckParse
//功能：校驗應用程序入口參數
//參數：argc等於main主函數argc參數，argv指向main主函數argv參數
//返回：若參數合法返回true，否則返回false
//備註：簡單的入口參數校驗
boolCheckParse(intargc,char**argv)
{
boolbRes=false;
if(argc!=4&&argc!=5)
gotoout;
if(argc==4&&argv[1][1]=='k')
{
//!生成公、私鑰對
if(!IsNumber(argv[2])||
!IsNumber(argv[3])||
atoi(argv[2])>MAX_PRIME||
atoi(argv[3])>MAX_PRIME)
gotoout;
}
elseif((argc==5)&&(argv[1][1]=='e'||argv[1][1]=='d'))
{
//!加密、解密操作
if(!IsNumber(argv[3])||
!IsNumber(argv[4])||
atoi(argv[3])>MAX_NUM||
atoi(argv[4])>MAX_NUM)
gotoout;
}
else
Usage(*argv);
bRes=true;
out:
returnbRes;
}
//名稱：kOption1
//功能：程序k選項操作：由素數P、Q生成私鑰d集合
//參數：uiP:程序入口參數P;uiQ:程序入口參數Q
//返回：執行正確返回生成私鑰數目，否則返回0
unsignedintkOption1(unsignedintuiP,unsignedintuiQ)
{
unsignedintuiRes=0;
if(!IsPrimeNumber(uiP))
{
printf("P輸入錯誤，P必須為(0,%d]素數",MAX_PRIME);
returnuiRes;
}
if(!IsPrimeNumber(uiQ))
{
printf("Q輸入錯誤，Q必須為(0,%d]素數",MAX_PRIME);
returnuiRes;
}
if(uiP==uiQ)
{
printf("素數P與素數Q相同，很容易根據公鑰n開平方得出素數P和Q，這種加密不安全，請更換素數! ");
returnuiRes;
}
printf("正在生成私鑰d集合...... ");
uiRes=MakePrivatedKeyd(uiP,uiQ);
returnuiRes;
}
//!程序主函數
intmain(intargc,char**argv)
{
unsignedintp,q,d,n,e;//twoprimep&q,publickey(n,e),privatekey(n,d)
CheckParse(argc,argv);
d=4828;//uid
if(argc==4)
{
p=atoi(argv[2]);
q=atoi(argv[3]);
MakePrivatedKeyd(p,q);
MakePairkey(p,q,d);
outputkey();
}
elseif(argc==5)
{
charFileName[20];
strcpy(FileName,argv[2]);
intlen;
if(argv[1][1]=='e')
{
unsignedchar*inBuffer=(unsignedchar*)malloc(MAX_FILE);//輸入緩沖區
int*cw=(int*)malloc(MAX_FILE);
len=FileIn(FileName,inBuffer);
e=atoi(argv[3]);
n=atoi(argv[4]);
rsa_encrypt(n,e,(char*)inBuffer,len,cw);
FileOut(cw,4*len,DECRYPT_FILE);
}
elseif(argv[1][1]=='d')
{
char*Buffer=(char*)malloc(MAX_FILE);//輸入緩沖區
int*cw=(int*)malloc(MAX_FILE);
len=FileIn(FileName,(unsignedchar*&)cw);
d=atoi(argv[3]);
n=atoi(argv[4]);
rsa_decrypt(n,d,cw,len,Buffer);
FileOut(Buffer,len/4,ENCRYPT_FILE);
}
}
return0;
}
❸ 求RSA演算法的源代碼（c語言）
這個是我幫個朋友寫的，寫的時候發現其實這個沒那麼復雜，不過，時間復雜度要高於那些成型了的，為人所熟知的rsa演算法的其他語言實現．
#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
:"
);/*輸入要加密的明文數字*/
scanf("%d\n",&m);
c=candp(m,e,n);
printf("so
,the
cipher
is
%4d",c);}
if(r==2)
{
printf("input
the
c
:"
);/*輸入要解密的密文數字*/
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');
}
}
❹ RSA加密解密演算法示例（C語言）
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <time.h>
#define PRIME_MAX 200 // 生成素數范圍
#define EXPONENT_MAX 200 // 生成指數e范圍
#define Element_Max 127 // 加密單元的最大值，這里為一個char, 即1Byte
char str_read[100]="hello world !"; // 待加密的原文
int str_encrypt[100]; // 加密後的內容
char str_decrypt[100]; // 解密出來的內容
int str_read_len; // str_read 的長度
int prime1, prime2; // 隨機生成的兩個質數
int mod, eular; // 模數和歐拉數
int pubKey, priKey; // 公鑰指數和私鑰指數
// 生成隨機素數，實際應用中，這兩個質數越大，就越難破解。
int randPrime()
{
int prime, prime2, i;
next:
prime = rand() % PRIME_MAX; // 隨機產生數
if (prime <= 1) goto next; // 不是質數，生成下一個隨機數
if (prime == 2 || prime == 3) return prime;
prime2 = prime / 2; // prime>=4, prime2 的平方必定大於 prime , 因此只檢查小於等於prime2的數
for (i = 2; i <= prime2; i++) // 判斷是否為素數
{
if (i * i > prime) return prime;
if (prime % i == 0) goto next; // 不是質數，生成下一個隨機數
}
}
// 歐幾里德演算法，判斷a,b互質
int gcd(int a, int b)
{
int temp;
while (b != 0) {
temp = b;
b = a % b;
a = temp;
}
return a;
}
//生成公鑰指數，條件是 1< e < 歐拉數，且與歐拉數互質。
int randExponent()
{
int e;
while (1)
{
e = rand() % eular; if (e < EXPONENT_MAX) break;
}
while (1)
{
if (gcd(e, eular) == 1) return e; e = (e + 1) % eular; if (e == 0 || e > EXPONENT_MAX) e = 2;
}
}
//生成私鑰指數
int inverse()
{
int d, x;
while (1)
{
d = rand() % eular;
x = pubKey * d % eular;
if (x == 1)
{
return d;
}
}
}
//加密函數
void jiami()
{
str_read_len = strlen(str_read); //從參數表示的地址往後找，找到第一個'\0'，即串尾。計算'\0'至首地址的「距離」，即隔了幾個字元，從而得出長度。
printf("密文是：");
for (int i = 0; i < str_read_len; i++)
{
int C = 1; int a = str_read[i], b = a % mod;
for (int j = 0; j < pubKey; j++) //實現加密
{
C = (C*b) % mod;
}
str_encrypt[i] = C;
printf("%d ", str_encrypt[i]);
}
printf("\n");
}
//解密函數
void jiemi()
{
int i=0; for (i = 0; i < str_read_len; i++)
{
int C = 1; int a = str_encrypt[i], b=a%mod;
for (int j = 0; j < priKey; j++)
{
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("隨機產生兩個素數：prime1 = %d , prime2 = %d ", prime1, prime2);
mod = prime1 * prime2; printf("模數：mod = prime1 * prime2 = %d \n", mod); if (mod > Element_Max) break; // 模數要大於每個加密單元的值
}
eular = (prime1 - 1) * (prime2 - 1); printf("歐拉數：eular=(prime1-1)*(prime2-1) = %d \n", eular);
pubKey = randExponent(); printf("公鑰指數：pubKey = %d\n", pubKey);
priKey = inverse(); printf("私鑰指數：priKey = %d\n私鑰為 (%d, %d)\n", priKey, priKey, mod);
jiami(); jiemi();
return 0;
}
❺ RSA加密演算法怎樣用C語言實現 急急急！！！
/*數據只能是大寫字母組成的字元串。
加密的時候，輸入Y，然後輸入要加密的文本（大寫字母）
解密的時候，輸入N，然後輸入一個整數n表示密文的個數，然後n個整數表示加密時候得到的密文。
*/
/*RSA algorithm */
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define MM 7081
#define KK 1789
#define PHIM 6912
#define PP 85
typedef char strtype[10000];
int len;
long nume[10000];
int change[126];
char antichange[37];
void initialize()
{ int i;
char c;
for (i = 11, c = 'A'; c <= 'Z'; c ++, i ++)
{ change[c] = i;
antichange[i] = c;
}
}
void changetonum(strtype str)
{ int l = strlen(str), i;
len = 0;
memset(nume, 0, sizeof(nume));
for (i = 0; i < l; i ++)
{ nume[len] = nume[len] * 100 + change[str[i]];
if (i % 2 == 1) len ++;
}
if (i % 2 != 0) len ++;
}
long binamod(long numb, long k)
{ if (k == 0) return 1;
long curr = binamod (numb, k / 2);
if (k % 2 == 0)
return curr * curr % MM;
else return (curr * curr) % MM * numb % MM;
}
long encode(long numb)
{ return binamod(numb, KK);
}
long decode(long numb)
{ return binamod(numb, PP);
}
main()
{ strtype str;
int i, a1, a2;
long curr;
initialize();
puts("Input 'Y' if encoding, otherwise input 'N':");
gets(str);
if (str[0] == 'Y')
{ gets(str);
changetonum(str);
printf("encoded: ");
for (i = 0; i < len; i ++)
{ if (i) putchar('-');
printf(" %ld ", encode(nume[i]));
}
putchar('\n');
}
else
{ scanf("%d", &len);
for (i = 0; i < len; i ++)
{ scanf("%ld", &curr);
curr = decode(curr);
a1 = curr / 100;
a2 = curr % 100;
printf("decoded: ");
if (a1 != 0) putchar(antichange[a1]);
if (a2 != 0) putchar(antichange[a2]);
}
putchar('\n');
}
putchar('\n');
system("PAUSE");
return 0;
}
/*
測試：
輸入：
Y
FERMAT
輸出：
encoded: 5192 - 2604 - 4222
輸入
N
3 5192 2604 4222
輸出
decoded: FERMAT
*/
❻ 求RSA加密解密演算法，c++源代碼
//下面程序由520huiqin編寫，已在VC++ 6.0下編譯通過
#include <iostream.h>
#include <math.h>
#include <stdio.h>
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;
};
/* 公鑰 */
struct PU {
Elemtype e;
Elemtype n;
} pu;
/* 私鑰 */
struct PR {
Elemtype d;
Elemtype n;
} pr;
/* 判定一個數是否為素數 */
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;
}
}
/* 將十進制數據轉化為二進制數組 */
void switch_to_bit(Elemtype b, Elemtype bin[32]) {
int n = 0;
while( b > 0) {
bin[n] = b % 2;
n++;
b /= 2;
}
}
/* 候選菜單，主界面 */
void Init() {
cout<<"*********************************************"<<endl;
cout<<"*** Welcome to use RSA encoder ***"<<endl;
cout<<"*** a.about ***"<<endl;
cout<<"*** e.encrypt ***"<<endl;
cout<<"*** d.decrypt ***"<<endl;
cout<<"*** s.setkey ***"<<endl;
cout<<"*** q.quit ***"<<endl;
cout<<"**********************************by*Terry***"<<endl;
cout<<"press a key:"<<endl;
}
/* 將兩個數排序，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;
}
/* 求最大公約數 */
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;
}
}
}
}
/* 用擴展的歐幾里得演算法求乘法逆元 */
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];
}
}
}
/* 快速模冪演算法 */
Elemtype molar_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;
}
/* 產生密鑰 */
void proce_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!"<<endl;
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 = "<<fn<<endl;
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: "<<pr.d<<" PR.n: "<<pr.n<<endl;
cout<<"PU.e: "<<pu.e<<" PU.n: "<<pu.n<<endl;
}
/* 加密 */
void encrypt() {
if(flag == 0) {
cout<<"setkey first:"<<endl;
proce_key();
}
cout<<"input m:";
cin>>m;
c = molar_multiplication(m,pu.e,pu.n);
cout<<"c is:"<<c<<endl;
}
/* 解密 */
void decrypt() {
if(flag == 0) {
cout<<"setkey first:"<<endl;
proce_key();
}
cout<<"input c:";
cin>>c;
m = molar_multiplication(c,pr.d,pr.n);
cout<<"m is:"<<m<<endl;
}
/* 版權信息 */
void about() {
cout<<"*********************************************"<<endl;
cout<<"*** by Terry ***"<<endl;
cout<<"*** right 2010,All rights reserved by ***"<<endl;
cout<<"*** Terry,technology supported by weizuo !***"<<endl;
cout<<"*** If you have any question, please mail ***"<<endl;
cout<<"*** to [email protected] ! ***"<<endl;
cout<<"*** Computer of science and engineering ***"<<endl;
cout<<"*** XiDian University 2010-4-29 ***"<<endl;
cout<<"*********************************************"<<endl;
cout<<endl<<endl;
Init();
}
/* 消息映射 */
MsgMap Messagemap[] = {
{'a',about},
{'s',proce_key},
{'d',decrypt},
{'e',encrypt},
{'q',NULL}
};
/* 主函數，提供循環 */
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++;
}
}
}
//歡迎分享，盜竊可恥
❼ 如何用C++實現RSA演算法
基礎
RSA演算法非常簡單,概述如下：
找兩素數p和q
取n=p*q
取t=(p-1)*(q-1)
取任何一個數e,要求滿足eperl -Mbigint -e "print 465**63%2773"
244
即用e對c解密後獲得m=244 , 該值和原始信息M相等.
字元串加密
把上面的過程集成一下我們就能實現一個對字元串加密解密的示例了.
每次取字元串中的一個字元的ascii值作為M進行計算,其輸出為加密後16進制
的數的字元串形式,按3位元組表示,如01F
代碼如下：
#!/usr/bin/perl -w
#RSA 計算過程學習程序編寫的測試程序
#watercloud 2003-8-12
#
use strict;
use Math::BigInt;
my %RSA_CORE = (n=>2773,e=>63,d=>847); #p=47,q=59
my $N=new Math::BigInt($RSA_CORE{n});
❽ C++實現RSA加密解密演算法
#include <iostream>
using namespace std;
template <class HugeInt>
HugeInt Power( const HugeInt & x, const HugeInt & n, // 求x^n mod p
const HugeInt & p )
{
if( n == 0 )
return 1;
HugeInt tmp = Power( ( x * x ) % p, n / 2, p );
if( n % 2 != 0 )
tmp = ( tmp * x ) % p;
return tmp;
}
template <class HugeInt>
void fullGcd( const HugeInt & a, const HugeInt & b, //
HugeInt & x, HugeInt & y )
{
HugeInt x1, y1;
if( b == 0 )
{
x = 1;
y = 0;
}
else
{
fullGcd( b, a % b, x1, y1 );
x = y1;
y = x1 - ( a / b ) * y1;
}
}
template <class HugeInt>
HugeInt inverse( const HugeInt & p, const HugeInt & q, // 求d
const HugeInt & e )
{
int fyn = ( 1 - p ) * ( 1 - q );
HugeInt x, y;
fullGcd( fyn, e, x, y );
return x > 0 ? x : x + e;
}
int main( )
{
cout << "Please input the plaintext: " << endl;
int m;
cin >> m;
cout << "Please input p,q and e: " << endl;
int p, q, e;
cin >> p >> q >> e;
int n = p * q;
int d = inverse( p, q, e );
int C = Power( m, e, n );
cout << "The ciphertext is: " << C << endl;
cout << "\n\nPlease input the ciphertext: " << endl;
cin >> C;
cout << "\n\nPlease input p, q and d: " << endl;
cin >> p >> q >> d;
n = p * q;
m = Power( C, d, n );
cout <<"The plaintext is: " << m << endl << endl;
system( "pause" );
return 0;
}
這就是RSA加密解密演算法