#include <string> #include <iostream> #include "time.h" #include <stdlib.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 = 七、/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<int(num/2);devider ) { if(num?vider==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((int)time(0)；//设置随机数种子，在rand以前调用一次 e=2 rand()%(t-3); return(e);//随机数 } //寻求最大的公共因素 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; } //判断最大公约数是否为1，如果是1，两个数字是互质的 ///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; i<model;) { over = over % model; if (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} //调用主函数 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; } // 随机选一个2<e<t ,使得gcd(t,e)=1 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;i<e1;i ) { b=b%m; b=b*tmp; } return 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; } ///重复平方求幂 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 //调用主函数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:"<<e<<endl; cout<<"d:"<<d<<endl; RSA_Encrytion( e, n);//加密 RSA_Decrytion( d,n);//解密 } int main() { RunRSA(); cout<<sizeof(unsigned int)<<endl; cout<<"原明文："<<msg<<endl; cout<<"密文："; for(int i=0;i<sizeof(msg);i ) { cout<<char(C_uint[i]); } cout<<endl; cout<<"解密明文："; for(int i=0;i<sizeof(msg);i ) { cout<<char(Dec[i]); } cout<<endl; system("pause"); return 0; } 参考自RSA加密解密C 实现_ermeiyao11的博客-CSDN博客_c 实现rsa加解密