C++ C+中高数值的模幂运算+;

C++ C+中高数值的模幂运算+;,c++,modulo,integer-overflow,exponentiation,C++,Modulo,Integer Overflow,Exponentiation,所以我最近一直在研究Miller-Rabin素性测试的实现。我把它限制在所有32位数字的范围内,因为这是一个有趣的项目,我正在做的是熟悉C++,我不想用任何64位一段时间。另外一个好处是,该算法对所有32位数字都是确定性的,因此我可以显著提高效率,因为我确切地知道要测试哪些证人 因此,对于较小的数字,该算法非常有效。然而,部分过程依赖于模幂运算,即(num^pow)%mod。那么比如说, 3 ^ 2 % 5 = 9 % 5 = 4 以下是我用于此模幂运算的代码: unsigned mod_

所以我最近一直在研究Miller-Rabin素性测试的实现。我把它限制在所有32位数字的范围内,因为这是一个有趣的项目,我正在做的是熟悉C++,我不想用任何64位一段时间。另外一个好处是,该算法对所有32位数字都是确定性的,因此我可以显著提高效率,因为我确切地知道要测试哪些证人

因此,对于较小的数字,该算法非常有效。然而,部分过程依赖于模幂运算,即(num^pow)%mod。那么比如说,

3 ^ 2 % 5 = 
9 % 5 = 
4
以下是我用于此模幂运算的代码:

unsigned mod_pow(unsigned num, unsigned pow, unsigned mod)
{
    unsigned test;
    for(test = 1; pow; pow >>= 1)
    {
        if (pow & 1)
            test = (test * num) % mod;
        num = (num * num) % mod;
    }

    return test;

}
正如您可能已经猜到的,当参数都是非常大的数字时,问题就会出现。例如,如果我想测试数字673109的素性,我必须在某一点上找到:

(2^168277)%673109

现在,2^168277是一个非常大的数字,在这个过程的某个地方它溢出了测试,这导致了错误的评估

另一方面,参数如

4000111222^3%1608

出于同样的原因,评估也不正确

有没有人对模幂运算有什么建议,可以防止这种溢出和/或操纵它来产生正确的结果?(在我看来,溢出只是模的另一种形式,即num%(UINT_MAX+1))

两件事:

  • 您是否使用了适当的数据类型?换句话说,UINT_MAX是否允许将673109作为参数
不,它没有,因为在某一点上,您的代码不起作用,因为在某一点上,您的
num=2^16
num=…
导致溢出。使用较大的数据类型保存此中间值

  • 在每一个可能的溢出机会取模怎么样,例如:

    test=((test%mod)*(num%mod))%mod

编辑:


您可以使用以下标识:

(a*b)(m型)==(a(m型))*(b(m型))(m型)

尝试直接使用它,并逐步改进

    if (pow & 1)
        test = ((test % mod) * (num % mod)) % mod;
    num = ((num % mod) * (num % mod)) % mod;
仍然适用于模幂运算。您的问题不是
2^168277
是一个非常大的数字,而是您的中间结果之一是一个相当大的数字(大于2^32),因为673109大于2^16

因此,我认为以下几点就可以了。我可能遗漏了一个细节,但基本思想是可行的,这就是“真实”密码如何实现大模幂运算(虽然不是32位和64位数字,而是使用永远不必大于2*log(模数)的bignum):

  • 如您所做的那样,从平方开始求幂
  • 在64位无符号整数中执行实际平方运算
  • 在每一步减少模673109,以回到32位范围内,就像您所做的那样
显然,如果C++实现没有64位整数,那就有点尴尬,尽管你总是可以伪造一个。 幻灯片22中有一个例子:,虽然它使用了非常小的数字(小于2^16),所以它可能无法说明您不知道的任何内容


另一个示例,
4000111222^3%1608
将在当前代码中工作,如果您在开始之前减少
4000111222
1608
<代码>1608 <代码>足够小,可以在32位INT/.P>>P>中安全地将任意两个MOD-1608号相乘。我最近为C++编写了一个RSA,有些混乱。

#include "BigInteger.h"
#include <iostream>
#include <sstream>
#include <stack>

BigInteger::BigInteger() {
    digits.push_back(0);
    negative = false;
}

BigInteger::~BigInteger() {
}

void BigInteger::addWithoutSign(BigInteger& c, const BigInteger& a, const BigInteger& b) {
    int sum_n_carry = 0;
    int n = (int)a.digits.size();
    if (n < (int)b.digits.size()) {
        n = b.digits.size();
    }
    c.digits.resize(n);
    for (int i = 0; i < n; ++i) {
        unsigned short a_digit = 0;
        unsigned short b_digit = 0;
        if (i < (int)a.digits.size()) {
            a_digit = a.digits[i];
        }
        if (i < (int)b.digits.size()) {
            b_digit = b.digits[i];
        }
        sum_n_carry += a_digit + b_digit;
        c.digits[i] = (sum_n_carry & 0xFFFF);
        sum_n_carry >>= 16;
    }
    if (sum_n_carry != 0) {
        putCarryInfront(c, sum_n_carry);
    }
    while (c.digits.size() > 1 && c.digits.back() == 0) {
        c.digits.pop_back();
    }
    //std::cout << a.toString() << " + " << b.toString() << " == " << c.toString() << std::endl;
}

void BigInteger::subWithoutSign(BigInteger& c, const BigInteger& a, const BigInteger& b) {
    int sub_n_borrow = 0;
    int n = a.digits.size();
    if (n < (int)b.digits.size())
        n = (int)b.digits.size();
    c.digits.resize(n);
    for (int i = 0; i < n; ++i) {
        unsigned short a_digit = 0;
        unsigned short b_digit = 0;
        if (i < (int)a.digits.size())
            a_digit = a.digits[i];
        if (i < (int)b.digits.size())
            b_digit = b.digits[i];
        sub_n_borrow += a_digit - b_digit;
        if (sub_n_borrow >= 0) {
            c.digits[i] = sub_n_borrow;
            sub_n_borrow = 0;
        } else {
            c.digits[i] = 0x10000 + sub_n_borrow;
            sub_n_borrow = -1;
        }
    }
    while (c.digits.size() > 1 && c.digits.back() == 0) {
        c.digits.pop_back();
    }
    //std::cout << a.toString() << " - " << b.toString() << " == " << c.toString() << std::endl;
}

int BigInteger::cmpWithoutSign(const BigInteger& a, const BigInteger& b) {
    int n = (int)a.digits.size();
    if (n < (int)b.digits.size())
        n = (int)b.digits.size();
    //std::cout << "cmp(" << a.toString() << ", " << b.toString() << ") == ";
    for (int i = n-1; i >= 0; --i) {
        unsigned short a_digit = 0;
        unsigned short b_digit = 0;
        if (i < (int)a.digits.size())
            a_digit = a.digits[i];
        if (i < (int)b.digits.size())
            b_digit = b.digits[i];
        if (a_digit < b_digit) {
            //std::cout << "-1" << std::endl;
            return -1;
        } else if (a_digit > b_digit) {
            //std::cout << "+1" << std::endl;
            return +1;
        }
    }
    //std::cout << "0" << std::endl;
    return 0;
}

void BigInteger::multByDigitWithoutSign(BigInteger& c, const BigInteger& a, unsigned short b) {
    unsigned int mult_n_carry = 0;
    c.digits.clear();
    c.digits.resize(a.digits.size());
    for (int i = 0; i < (int)a.digits.size(); ++i) {
        unsigned short a_digit = 0;
        unsigned short b_digit = b;
        if (i < (int)a.digits.size())
            a_digit = a.digits[i];
        mult_n_carry += a_digit * b_digit;
        c.digits[i] = (mult_n_carry & 0xFFFF);
        mult_n_carry >>= 16;
    }
    if (mult_n_carry != 0) {
        putCarryInfront(c, mult_n_carry);
    }
    //std::cout << a.toString() << " x " << b << " == " << c.toString() << std::endl;
}

void BigInteger::shiftLeftByBase(BigInteger& b, const BigInteger& a, int times) {
    b.digits.resize(a.digits.size() + times);
    for (int i = 0; i < times; ++i) {
        b.digits[i] = 0;
    }
    for (int i = 0; i < (int)a.digits.size(); ++i) {
        b.digits[i + times] = a.digits[i];
    }
}

void BigInteger::shiftRight(BigInteger& a) {
    //std::cout << "shr " << a.toString() << " == ";
    for (int i = 0; i < (int)a.digits.size(); ++i) {
        a.digits[i] >>= 1;
        if (i+1 < (int)a.digits.size()) {
            if ((a.digits[i+1] & 0x1) != 0) {
                a.digits[i] |= 0x8000;
            }
        }
    }
    //std::cout << a.toString() << std::endl;
}

void BigInteger::shiftLeft(BigInteger& a) {
    bool lastBit = false;
    for (int i = 0; i < (int)a.digits.size(); ++i) {
        bool bit = (a.digits[i] & 0x8000) != 0;
        a.digits[i] <<= 1;
        if (lastBit)
            a.digits[i] |= 1;
        lastBit = bit;
    }
    if (lastBit) {
        a.digits.push_back(1);
    }
}

void BigInteger::putCarryInfront(BigInteger& a, unsigned short carry) {
    BigInteger b;
    b.negative = a.negative;
    b.digits.resize(a.digits.size() + 1);
    b.digits[a.digits.size()] = carry;
    for (int i = 0; i < (int)a.digits.size(); ++i) {
        b.digits[i] = a.digits[i];
    }
    a.digits.swap(b.digits);
}

void BigInteger::divideWithoutSign(BigInteger& c, BigInteger& d, const BigInteger& a, const BigInteger& b) {
    c.digits.clear();
    c.digits.push_back(0);
    BigInteger two("2");
    BigInteger e = b;
    BigInteger f("1");
    BigInteger g = a;
    BigInteger one("1");
    while (cmpWithoutSign(g, e) >= 0) {
        shiftLeft(e);
        shiftLeft(f);
    }
    shiftRight(e);
    shiftRight(f);
    while (cmpWithoutSign(g, b) >= 0) {
        g -= e;
        c += f;
        while (cmpWithoutSign(g, e) < 0) {
            shiftRight(e);
            shiftRight(f);
        }
    }
    e = c;
    e *= b;
    f = a;
    f -= e;
    d = f;
}

BigInteger::BigInteger(const BigInteger& other) {
    digits = other.digits;
    negative = other.negative;
}

BigInteger::BigInteger(const char* other) {
    digits.push_back(0);
    negative = false;
    BigInteger ten;
    ten.digits[0] = 10;
    const char* c = other;
    bool make_negative = false;
    if (*c == '-') {
        make_negative = true;
        ++c;
    }
    while (*c != 0) {
        BigInteger digit;
        digit.digits[0] = *c - '0';
        *this *= ten;
        *this += digit;
        ++c;
    }
    negative = make_negative;
}

bool BigInteger::isOdd() const {
    return (digits[0] & 0x1) != 0;
}

BigInteger& BigInteger::operator=(const BigInteger& other) {
    if (this == &other) // handle self assignment
        return *this;
    digits = other.digits;
    negative = other.negative;
    return *this;
}

BigInteger& BigInteger::operator+=(const BigInteger& other) {
    BigInteger result;
    if (negative) {
        if (other.negative) {
            result.negative = true;
            addWithoutSign(result, *this, other);
        } else {
            int a = cmpWithoutSign(*this, other);
            if (a < 0) {
                result.negative = false;
                subWithoutSign(result, other, *this);
            } else if (a > 0) {
                result.negative = true;
                subWithoutSign(result, *this, other);
            } else {
                result.negative = false;
                result.digits.clear();
                result.digits.push_back(0);
            }
        }
    } else {
        if (other.negative) {
            int a = cmpWithoutSign(*this, other);
            if (a < 0) {
                result.negative = true;
                subWithoutSign(result, other, *this);
            } else if (a > 0) {
                result.negative = false;
                subWithoutSign(result, *this, other);
            } else {
                result.negative = false;
                result.digits.clear();
                result.digits.push_back(0);
            }
        } else {
            result.negative = false;
            addWithoutSign(result, *this, other);
        }
    }
    negative = result.negative;
    digits.swap(result.digits);
    return *this;
}

BigInteger& BigInteger::operator-=(const BigInteger& other) {
    BigInteger neg_other = other;
    neg_other.negative = !neg_other.negative;
    return *this += neg_other;
}

BigInteger& BigInteger::operator*=(const BigInteger& other) {
    BigInteger result;
    for (int i = 0; i < (int)digits.size(); ++i) {
        BigInteger mult;
        multByDigitWithoutSign(mult, other, digits[i]);
        BigInteger shift;
        shiftLeftByBase(shift, mult, i);
        BigInteger add;
        addWithoutSign(add, result, shift);
        result = add;
    }
    if (negative != other.negative) {
        result.negative = true;
    } else {
        result.negative = false;
    }
    //std::cout << toString() << " x " << other.toString() << " == " << result.toString() << std::endl;
    negative = result.negative;
    digits.swap(result.digits);
    return *this;
}

BigInteger& BigInteger::operator/=(const BigInteger& other) {
    BigInteger result, tmp;
    divideWithoutSign(result, tmp, *this, other);
    result.negative = (negative != other.negative);
    negative = result.negative;
    digits.swap(result.digits);
    return *this;
}

BigInteger& BigInteger::operator%=(const BigInteger& other) {
    BigInteger c, d;
    divideWithoutSign(c, d, *this, other);
    *this = d;
    return *this;
}

bool BigInteger::operator>(const BigInteger& other) const {
    if (negative) {
        if (other.negative) {
            return cmpWithoutSign(*this, other) < 0;
        } else {
            return false;
        }
    } else {
        if (other.negative) {
            return true;
        } else {
            return cmpWithoutSign(*this, other) > 0;
        }
    }
}

BigInteger& BigInteger::powAssignUnderMod(const BigInteger& exponent, const BigInteger& modulus) {
    BigInteger zero("0");
    BigInteger one("1");
    BigInteger e = exponent;
    BigInteger base = *this;
    *this = one;
    while (cmpWithoutSign(e, zero) != 0) {
        //std::cout << e.toString() << " : " << toString() << " : " << base.toString() << std::endl;
        if (e.isOdd()) {
            *this *= base;
            *this %= modulus;
        }
        shiftRight(e);
        base *= BigInteger(base);
        base %= modulus;
    }
    return *this;
}

std::string BigInteger::toString() const {
    std::ostringstream os;
    if (negative)
        os << "-";
    BigInteger tmp = *this;
    BigInteger zero("0");
    BigInteger ten("10");
    tmp.negative = false;
    std::stack<char> s;
    while (cmpWithoutSign(tmp, zero) != 0) {
        BigInteger tmp2, tmp3;
        divideWithoutSign(tmp2, tmp3, tmp, ten);
        s.push((char)(tmp3.digits[0] + '0'));
        tmp = tmp2;
    }
    while (!s.empty()) {
        os << s.top();
        s.pop();
    }
    /*
    for (int i = digits.size()-1; i >= 0; --i) {
        os << digits[i];
        if (i != 0) {
            os << ",";
        }
    }
    */
    return os.str();

它的速度也很快,并且有无限位数。

LL
代表
long-long-int

    package playTime;

    public class play {

        public static long count = 0; 
        public static long binSlots = 10; 
        public static long y = 645; 
        public static long finalValue = 1; 
        public static long x = 11; 

        public static void main(String[] args){

            int[] binArray = new int[]{0,0,1,0,0,0,0,1,0,1};  

            x = BME(x, count, binArray); 

            System.out.print("\nfinal value:"+finalValue);

        }

        public static long BME(long x, long count, int[] binArray){

            if(count == binSlots){
                return finalValue; 
            }

            if(binArray[(int) count] == 1){
                finalValue = finalValue*x%y; 
            }

            x = (x*x)%y; 
            System.out.print("Array("+binArray[(int) count]+") "
                            +"x("+x+")" +" finalVal("+              finalValue + ")\n");

            count++; 


            return BME(x, count,binArray); 
        }

    }
LL power_mod(LL a, LL k) {
    if (k == 0)
        return 1;
    LL temp = power(a, k/2);
    LL res;

    res = ( ( temp % P ) * (temp % P) ) % P;
    if (k % 2 == 1)
        res = ((a % P) * (res % P)) % P;
    return res;
}
使用上述递归函数查找数字的mod exp。这不会导致溢出,因为它是以自下而上的方式计算的

以下各项的样本测试运行:
a=2
k=168277
显示输出为518358,这是正确的,函数在
O(log(k))
时间内运行

谢谢你们两位的建议,但是由于算法的性质,test和num总是小于mod,所以:{(test%mod)=test}和{(num%mod)=test}因此标识不能帮助我,因为即使num和test小于mod,函数也会失败。另外,无符号整数允许我使用673109作为参数。UINT_MAX=4294967295为我的电脑。谢谢你,伙计,这就成功了。出于好奇,你知道有什么方法不需要使用更大的内存吗?我相信它们会派上用场的。据我所知不是这样。您需要将两个数字相乘到673108,mod 673109。显然,你可以把事情分解,用更小的“数字”做长乘法,比如说2^10。但是,一旦在软件中实现乘除运算,您就可以实现它,以实现两个32位值相乘得到64位结果的特殊情况,然后进行除法以提取32位余数。可能有一些硬核心优化,你需要的最小限度,但我不知道他们,伪造64位int在C++中并不难。这是我用java写的代码非常快。我使用的例子是11^644MOD6451.我们知道645的二进制数是10100000100。输出是数组(0)x(121)finalVal(1)数组(0)x(451)finalVal(1)数组(1)x(226)finalVal(451)数组(0)x(121)finalVal(451)数组(0)x(451)finalVal(451)数组(0)x(226)finalVal(451)数组(0)x(121)finalVal(451)数组(1)finalVal(391)数组(0)x(226)finalVal(391)x(121)最终(1)最终值:感谢分享!一个问题,数字是std::vector吗?是的,但在机罩下的65536基地工作,而不是10基地。
    package playTime;

    public class play {

        public static long count = 0; 
        public static long binSlots = 10; 
        public static long y = 645; 
        public static long finalValue = 1; 
        public static long x = 11; 

        public static void main(String[] args){

            int[] binArray = new int[]{0,0,1,0,0,0,0,1,0,1};  

            x = BME(x, count, binArray); 

            System.out.print("\nfinal value:"+finalValue);

        }

        public static long BME(long x, long count, int[] binArray){

            if(count == binSlots){
                return finalValue; 
            }

            if(binArray[(int) count] == 1){
                finalValue = finalValue*x%y; 
            }

            x = (x*x)%y; 
            System.out.print("Array("+binArray[(int) count]+") "
                            +"x("+x+")" +" finalVal("+              finalValue + ")\n");

            count++; 


            return BME(x, count,binArray); 
        }

    }
LL power_mod(LL a, LL k) {
    if (k == 0)
        return 1;
    LL temp = power(a, k/2);
    LL res;

    res = ( ( temp % P ) * (temp % P) ) % P;
    if (k % 2 == 1)
        res = ((a % P) * (res % P)) % P;
    return res;
}