RSA加密解密C++实现

2019-04-14 16:05发布

站内文章 / 模拟电子
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实现过程：
1 随意选择两个大的质数p和q，p不等于q，计算N=p*q。
2 根据欧拉函数，求得r = (p-1)(q-1)
3 选择一个小于 r 的整数 e，求得 e 关于模 r 的模反元素，命名为d。（模反元素存在，当且仅当e与r互质）
将 p 和 q 的记录销毁。 -大数模幂运算快速算法
参考网址 代码： #include #include #include "time.h" using namespace std; unsigned char msg[8] ={ 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h' };//明文 unsigned int C_uint[10]; unsigned int Dec[10]; unsigned int PP = 7;//两个素数 unsigned int QQ = 17; //判断p是否为素数 //在产生密钥函数中调用 bool JudgePrimeNum(unsigned int num) { unsigned int devider=2; for(;devider<unsigned int(num/2);devider++) { if(num%devider==0) return false; } return true; } //产生2到t-1的随机数e //在产生密钥函数中调用 unsigned int RandomlyGenerateE(unsigned int t) { unsigned int e=0; srand((unsigned int)time(0));//设置随机数种子，在rand之前调用一次即可 e=2+rand()%(t-3); return(e);//随机数 } //求最大公因素 void Gcd(unsigned int BigNum,unsigned int SmallNum,unsigned int &MaxGcd ) { int tmp=0; while(BigNum%SmallNum) { tmp=SmallNum; SmallNum=BigNum%SmallNum; BigNum=tmp; } MaxGcd=SmallNum; } //判断最大公约数数是否是1，是1的话两个数就互质 //在产生密钥函数中调用 bool JudgeGcd_1(unsigned int BigNum,unsigned int SmallNum) { unsigned int M=0; Gcd( BigNum,SmallNum,M); if(M==1) return true; else return false; } //　3*7 = 21; 21 % 20 = 1 ; 所以3,7 互为 20 的 模逆. //　9*3 = 27; 27 % 26 = 1 ; 所以9,3 互为 26 的 模逆. //==利用加的模等于模的加求e*d = 1 mod model 中的d //在产生密钥函数中调用 int Moni(unsigned int e, unsigned int model, unsigned int* d) { unsigned int i; unsigned int over = e; for (i = 1; iif (over == 1) { *d = i; return 1; } else { if (over + e <= model) { do { i++; over += e; } while (over + e <= model); } else { i++; over += e; } } } return 0; } //产生密钥函数 其中p q互异至数 ，公钥P{e,n)，私钥S{d,n} //在主函数调用 void ProduceKey(unsigned int p,unsigned int q,unsigned int &e,unsigned int &d,unsigned int &n) { unsigned int t=0; while(!JudgePrimeNum(p)) { cout<<"p不是质数，请重新输入p:"; cin>>p; } while(!JudgePrimeNum(q)) { cout<<"q不是质数，请重新输入q:"; cin>>q; } // 随机选择一个2 n=p*q; t=(p-1)*(q-1); e=RandomlyGenerateE(t); while(!JudgeGcd_1(t,e)) { e=RandomlyGenerateE(t); } //计算d,使得e*d=1 mod t Moni(e,t,&d); } unsigned int Modular_Ex(unsigned int e1,int b,const unsigned int m) { unsigned int i; unsigned int tmp=b; for(i=0;ireturn b/tmp; } //二进制转换 int BianaryTransform(int num, int bin_num[]) { int i = 0, mod = 0; //转换为二进制，逆向暂存temp[]数组中 while(num != 0) { mod = num%2; bin_num[i] = mod; num = num/2; i++; } //返回二进制数的位数 return i; } //反复平方求幂 unsigned int Modular_Exonentiation(unsigned int a, int b, int n) { int c = 0, bin_num[1000]; long long d = 1; int k = BianaryTransform(b, bin_num)-1; for(int i = k; i >= 0; i--) { c = 2*c; d = (d*d)%n; if(bin_num[i] == 1) { c = c + 1; d = (d*a)%n; } } return d; } //RSA加密 //在主函数调用a*a%n =((a%n)*a)%n void RSA_Encrytion(unsigned int e1,const unsigned int n1) { unsigned int i; unsigned int tmp; int j; for(j=0;j<sizeof(msg);j++) { C_uint[j]=Modular_Exonentiation(msg[j],e1,n1); } } //RSA解密 void RSA_Decrytion(unsigned int d2,const unsigned int n2) { unsigned int i; unsigned int tmp; int j; for(j=0;j<sizeof(msg);j++) { Dec[j]=Modular_Exonentiation(C_uint[j],d2,n2); } } void RunRSA() { unsigned int e=0; unsigned int d=0; unsigned int n=0;//n=p*q ProduceKey(PP,QQ,e,d,n); cout<<"e:"<cout<<"d:"<//加密 RSA_Decrytion( d,n);//解密 } int main() { RunRSA(); cout<<sizeof(unsigned int); cout<<"原明文："<cout<<"密文："<for(int i=0;i<sizeof(msg);i++) { cout<<char(C_uint[i]); } cout<cout<<"解密得到的明文："<for(int i=0;i<sizeof(msg);i++) { cout<<char(Dec[i]); } system("pause"); return 0; }

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