我写的RSA简易算法，为啥结果错误？

[zwdnet]

为了理解RSA算法，我按照网上的资料（http://www.winu.cn/space-14160-do-blog-id-11182.html）用C++写了个RSA加密解密的程序，为了理解算法本身，我没有用大整数。编译通过后加密解密测试出现错误结果，加密整数20结果密文和明文都是0，麻烦大家帮忙看看是我哪里理解有误。谢谢。

#include <iostream>
#include <cmath>
#include <cstdlib>
#include <ctime>

using namespace std;

//判断数字是否为素数
bool IsPrime(long N)
{
    if (N == 2 || N == 3 ||
        N == 5 || N == 7)
    {
        return true;
    }
    if (N % 2 == 0 || N % 3 == 0 ||
        N % 5 == 0 || N % 7 == 0)
    {
        return false;
    }
    long sqrtN = long(sqrt(N)) + 1;
    for (long i = 3; i < sqrtN; i += 2)
    {
        if (N % i == 0)
        {
            return false;
        }
    }
    return true;
}

//查找大于a,小于b范围内的一个随机素数
long FindPrime(long a, long b)
{
    long N = -1;
    long n = b - a + 1;
    srand(time(NULL));
    do
    {
        N = rand() % n + a;
    }while (!IsPrime(N));
    return N;
}


//求m,n的最大公约数
long gcd(long m, long n)
{
    if (m < n)
    {
        long tmp = m;
        m = n;
        n = tmp;
    }
    if (n == 0)
        return m;
    else
        return gcd(n, m%n);
}

//求a的n次方
long mypow(long a, long n)
{
    long tmp = a;
    for (int i = 2; i <= n; i++)
    {
        tmp = tmp * a;
    }
    return tmp;
}

int main()
{
    long a = 10, b = 20;
    long p = FindPrime(a, b);
    long q = FindPrime(a, b);
    while (p == q)
    {
        q = FindPrime(a, b);
    }
    long R = (p-1)*(q-1);
    //查找公钥r
    long r = 2;
    do
    {
        r++;
    }while (gcd(r, R) != 1);
    cout<<"p = "<<p<<endl
        <<"q = "<<q<<endl
        <<"r = "<<r<<endl;
    //查找密钥m
    long m = 2;
    for (m = 2; m <= R; m++)
    {
        if ((r*m) % R == 1)
        {
            break;
        }
    }
    cout<<"R = "<<R<<endl;
    cout<<"m = "<<m<<endl;
    long n = p*q;
    cout<<"n = "<<n<<endl;
    //加密解密测试
    long M = 20; //要加密的内容
    cout<<"明文 = "<<M<<endl;
    long Mm = mypow(M, m)%n;
    cout<<"密文 = "<<Mm<<endl;
    long Dm = mypow(Mm, r)%n;
    cout<<"解密 = "<<Dm<<endl;
    return 0;
}

程序的一个输出为:
p = 17
q = 11
r = 3
R = 160
m = 107
n = 187
明文 = 20
密文 = 0
解密 = 0

Process returned 0 (0x0) execution time : 2.656 s
Press any key to continue.

[butwang]

long Mm = mypow(M, m)%n;

Mm = (20的107次方)%187,数据很大溢出了，数据被截断为0了

[zwdnet]

谢谢，是数据溢出的问题。我把p,q的范围改成[5,10],再把待加密数据改成2，就测试成功了。