RSA算法的C++string实现(模幂算法和欧几里得算法的使用)后附思路

void resetNumA(string numAStr);
//使用string重置numB
void resetNumB(string numBStr);
//将数组转换为字符串，用于输出
string getNumString(int* num);
//判断两个数字哪个大
int compare(string numAStr, string numBStr);
//加法
string sum(string numAStr, string numBStr);
//减法
string sub(string numAStr, string numBStr);
//乘法
string mul(string numAStr, string numBStr);
//除
string div(string numAStr, string numBStr);
//取余
string mod(string numAStr, string numBStr);
//模幂算法
string getMod(string m, string pow, string n);
//求2的n次方函数
string mul_2(int i);
//求m的n次方
string mul_m(string m, string n);
#endif

#include<iostream>
#include"operation.h"
#include<string>
#include<vector>
using namespace std;

//结果支持的最大位数
const static int M = 2000;


int numA[M];
int numB[M];

//使用string重置numA
void resetNumA(string numAStr)
{
    memset(numA, 0, M * sizeof(int));

    //将字符串的每一位都转换成数字传入数组
    for (int i = 0; i < numAStr.length(); i++)
    {
        numA[i] = numAStr[numAStr.length() - i - 1] - '0';
    }
}

//使用string重置numB
void resetNumB(string numBStr)
{
    memset(numB, 0, M * sizeof(int));

    //将字符串的每一位都转换成数字传入数组
    for (int i = 0; i < numBStr.length(); i++)
    {
        numB[i] = numBStr[numBStr.length() - i - 1] - '0';
    }
}


//将数组转换为字符串，用于输出
string getNumString(int* num)
{
    string numString;
    bool isBegin = false;
    for (int i = M - 1; i >= 0; i--)
    {
        if (num[i] != 0)
        {
            isBegin = true;
        }

        if (isBegin)
        {
            numString += num[i] + '0';
        }
    }
    return numString;
}

//判断两个数字哪个大
int compare(string numAStr, string numBStr)
{
    int i = 0;
    int la = 0;
    while (numAStr[i] == '0') {
        la++;
        i++;
    }
    i = 0;
    int lb = 0;
    while (numBStr[i] == '0') {
        lb++;
        i++;
    }
    string a(numAStr.substr(la, numAStr.length()));
    string b(numBStr.substr(lb, numBStr.length()));
    if (a.length() > b.length())
    {
        return 1;
    }
    else if (a.length() < b.length())
    {
        return -1;
    }
    else
    {
        for (int i = 0; i < a.length(); i++)
        {
            if (a[i]>b[i])
            {
                return 1;
            }

            if (a[i]<b[i])
            {
                return -1;
            }
        }
        return 0;
    }
}

//加法
string sum(string numAStr, string numBStr)
{
    resetNumA(numAStr);
    resetNumB(numBStr);

    for (int i = 0; i < M; i++)
    {
        //结果保存在numA中
        numA[i] += numB[i];

        //数字大于9则进位
        if (numA[i]>9)
        {
            numA[i] -= 10;
            numA[i + 1]++;
        }
    }

    return getNumString(numA);
}

//减法
string sub(string numAStr, string numBStr)
{
    bool isNegative = false;

    //如果numA比numB小
    //则结果为负数
    //调换位置进行计算
    if (compare(numAStr, numBStr) == -1)
    {
        isNegative = true;
        string temp = numAStr;
        numAStr = numBStr;
        numBStr = temp;
    }
    else if (compare(numAStr, numBStr) == 0)
    {
        return "0";
    }

    resetNumA(numAStr);
    resetNumB(numBStr);



    for (int i = 0; i < M; i++)
    {
        //减数小于被减数就借位
        if (numA[i]<numB[i])
        {
            numA[i] = numA[i] + 10 - numB[i];
            numA[i + 1]--;
        }
        else
        {
            numA[i] -= numB[i];
        }
    }
    if (isNegative)
    {
        return "-" + getNumString(numA);
    }
    else
    {
        return getNumString(numA);
    }

}

//乘法

string mul(string numAStr, string numBStr)
{
    resetNumA(numAStr);
    resetNumB(numBStr);

    vector<string> nums;
    for (int i = 0; i < numBStr.length(); i++)
    {
        //初始化一个临时数据来保存被乘数与乘数的某一位相乘的结果
        int temp[M];
        memset(temp, 0, M * sizeof(int));


        for (int j = i; j < numAStr.length() + i; j++)
        {
            temp[j] += numA[j - i] * numB[i] % 10;

            temp[j + 1] = numA[j - i] * numB[i] / 10;

            //如果大于9，那么就做进位处理
            if (temp[j]>9)
            {
                temp[j] -= 10;
                temp[j + 1]++;
            }
        }
        nums.push_back(getNumString(temp));
    }

    //每位相乘的结果再用加法加起来
    string result = nums[0];
    for (int i = 1; i < nums.size(); i++)
    {
        result = sum(result, nums[i]);
    }

    return result;
}



//除,结果精确到个位
string div(string numAStr, string numBStr)
{
    resetNumA(numAStr);
    resetNumB(numBStr);

    string result;
    string left;

    if (compare(numAStr, numBStr) == -1)
    {
        return "0";
    }

    //标记第一个不为0的位数的出现
    bool flag = false;
    for (int i = 0; i < numAStr.length(); i++)
    {
        left += numAStr[i];

        //余数比除数大
        if (compare(left, numBStr) == 1)
        {
            flag = true;

            int count = 1;
            string temp = numBStr;

            while (true)
            {
                //每循环一次加上一个余数
                temp = sum(temp, numBStr);

                //余数仍然大于除数，继续累加
                if (compare(left, temp) == 1)
                {
                    count++;
                }
                //余数小于除数
                //可以计算结果
                else if (compare(left, temp) == -1)
                {
                    result += count + '0';
                    left = sub(left, sub(temp, numBStr));
                    break;
                }
                //此时余数刚好是除数的倍数
                else if (compare(left, temp) == 0)
                {
                    count++;
                    result += count + '0';
                    left = "";
                    break;
                }
            }
        }
        //刚好除尽
        else if (compare(left, numBStr) == 0)
        {
            flag = true;
            result += "1";
            left = "";
        }
        //余数比除数小，跳到下一位
        else if (flag)
        {
            result += "0";
        }


    }

    return result;
}
//取模
string mod(string numAStr, string numBStr)
{
    
    string result = "0";

    if (compare(numAStr, numBStr) == -1)
    {
        return numAStr;
    }
    else if (compare(numAStr, numBStr) == 0) {
        return result;
    }
    else {
        string d = div(numAStr, numBStr);
        string x = mul(numBStr, d);
        result = sub(numAStr, x);
        return result;
    }
}

//加密解密模幂算法
string getMod(string m, string pow, string n)
{
    string temp;
    while (compare(pow,"1") == 1) {
        if(mod(pow,"2")=="1")
            temp = temp + "1";
        else
            temp = temp + "0";
        pow = div(pow,"2");
    }
    temp = temp + "1";
    int length = temp.length();
    string T[M];
    string M = "1";
    T[0] = mod(m, n);
    for (int i = 1; i < length; i++) {
        T[i] = mod(mul(T[i-1],T[i-1]),n);
    }
    for (int i = 0; i < length; i++) {
        if (temp[i] == '1') {
            M = mul(M, T[i]);
        }
    }
    string result = mod(M, n);
    return result;
}

//求2的n次方函数
string mul_2(int i) {
    string result = "2";
    for (int j = 1; j < i; j++) {
        result = mul(result, "2");
    }
    if (i == 0)
        return "1";
    else
        return result;
}

//求m的t次方函数
string mul_m(string m,string t) {
    string result = m;
    string j;
    for ( j = "1"; compare(j, t) == -1; j = sum(j, "1")) {
        result = mul(result, m);
    }
        return result;
}

#include<iostream>
#include"operation.h"
using namespace std;


int main() {
    string char_number_p;
    string char_number_q;

    string temp[3000][4];
    //p=17,q=11
    //p=17,q=19
    //p=41,q=43
    //67,71
    //797 809
    //49993 49999
    //116747 110221
    //1000017077
    //141008192341187
    char_number_p="100010001707701";
    char_number_q="141008192341187";

    string n = mul(char_number_p, char_number_q); 
    string tmp1 = "1";
    string On = mul(sub(char_number_p, tmp1),sub(char_number_q, tmp1));
    
    string e = "65537";

    //拓展欧几里得算法
    temp[0][0] = e;
    temp[0][1] = "";
    temp[0][2] = On;
    temp[0][3] = "";
    int i = 1;
    do {
        temp[i][0] = temp[i - 1][0];
        temp[i][1] = "";
        temp[i][2] = temp[i - 1][2];
        temp[i][3] = "";
        if (compare(temp[i][0], temp[i][2]) == 1) {
            temp[i][0] = mod(temp[i][0], temp[i][2]);
        }
        else {
            temp[i][2] = mod(temp[i][2], temp[i][0]);
        }
        i++;
    } while (compare(temp[i - 1][0], "1") == 1);
    int value = i - 1;
    temp[value][1] = "1";
    value--;
    while (value >= 0) {
        if (temp[value + 1][1] != "")
            temp[value][3] = div(sub(mul(temp[value][0], temp[value + 1][1]), "1"), temp[value][2]);
        else if (temp[value + 1][3] != "")
            temp[value][1] = div(sum(mul(temp[value][2], temp[value + 1][3]), "1"), temp[value][0]);
        value--;
    }
    string d = temp[0][1];

    cout << "e:" << e << endl;
    cout << "p*q:" << n << endl;
    cout << "O(n):" << On << endl;
    cout << "d:" << d << endl;
    cout << "密钥{ " << e << " , " << n << " }" << endl;
    cout << "公钥{ " << d << " , " << char_number_p <<" , "<< char_number_q << " }" << endl;

    string miwen;
    cout << "请输入密文(暂小于10的29次方)" << endl;
    cin >> miwen;
    cout << "密文:" << miwen << endl;
    string C1 = getMod(miwen, e, n);
    cout << "加密后的明文:" << C1 << endl;
    string M1 = getMod(C1, d, n);
    cout << "解密后的密文:" << M1 << endl;

    cout << "请输入密文(暂小于10的29次方)" << endl;
    cin >> miwen;
    string C2 = getMod(miwen, e, n);
    cout << "加密后的明文:" << C2 << endl;
    string M2 = getMod(C2, d, n);
    cout << "解密后的密文:" << M2 << endl;

    cout << "请输入密文(暂小于10的29次方)" << endl;
    cin >> miwen;
    string C3 = getMod(miwen, e, n);
    cout << "加密后的明文:" << C3 << endl;
    string M3 = getMod(C3, d, n);
    cout << "解密后的密文:" << M3 << endl;
    
    system("pause");
    return 0;
}

以上为测试实现的代码。

思路：先实现string类型的加减乘除的重写，以及用到的其他的操作函数，比较，取模，加解密函数，以及其他的m的n次方的操作。

然后在主函数里面嵌入欧几里得算法，调用求得d，之后就是简单的调用加解密函数

pq的值也是测试得到的

这是运行截图：

//拓展欧几里得算法
    temp[0][0] = e;
    temp[0][1] = "";
    temp[0][2] = On;
    temp[0][3] = "";
    int i = 1;
    do {
        temp[i][0] = temp[i - 1][0];
        temp[i][1] = "";
        temp[i][2] = temp[i - 1][2];
        temp[i][3] = "";
        if (compare(temp[i][0], temp[i][2]) == 1) {
            temp[i][0] = mod(temp[i][0], temp[i][2]);
        }
        else {
            temp[i][2] = mod(temp[i][2], temp[i][0]);
        }
        i++;
    } while (compare(temp[i - 1][0], "1") == 1);
    int value = i - 1;
    temp[value][1] = "1";
    value--;
    while (value >= 0) {
        if (temp[value + 1][1] != "")
            temp[value][3] = div(sub(mul(temp[value][0], temp[value + 1][1]), "1"), temp[value][2]);
        else if (temp[value + 1][3] != "")
            temp[value][1] = div(sum(mul(temp[value][2], temp[value + 1][3]), "1"), temp[value][0]);
        value--;
    }
    string d = temp[0][1];
这部分拓展的欧几里得算法也可以单独使用，这里是用空间换取的时间效率。