ELGamal数字签名方案的实现
1． 问题描述
为简化问题，我们取p=19,g=2,私钥x=9,则公钥y=29 mod 19=18。消息m的ELGamal签名为(r,s),其中r=gk mod p,s=(h(m)-xr)k-1 mod (p-1)
2．基本要求
考虑p取大素数的情况。
3． 实现提示
① 模n求逆a-1modn运算。
② 模n的大数幂乘运算
由于大素数的本原元要求得很费事,所以签名所需要的数值我已经事先给出，当然这些数值比较小，有兴趣的同学可以自行将数值变大.
ELGamal离不开大数包的支持！
BigInteger.h
#pragma once
#include <cstring>
#include <string>
#include <algorithm>
#include <assert.h>
#include <ctime>
#include <iostream>
using namespace std;
const int maxLength = 512;
const int primeLength = 30;
const int primesBelow2000[303] = {
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97,
101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199,
211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293,
307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397,
401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499,
503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599,
601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691,
701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797,
809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887,
907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997,
1009, 1013, 1019, 1021, 1031, 1033, 1039, 1049, 1051, 1061, 1063, 1069, 1087, 1091, 1093, 1097,
1103, 1109, 1117, 1123, 1129, 1151, 1153, 1163, 1171, 1181, 1187, 1193,
1201, 1213, 1217, 1223, 1229, 1231, 1237, 1249, 1259, 1277, 1279, 1283, 1289, 1291, 1297,
1301, 1303, 1307, 1319, 1321, 1327, 1361, 1367, 1373, 1381, 1399,
1409, 1423, 1427, 1429, 1433, 1439, 1447, 1451, 1453, 1459, 1471, 1481, 1483, 1487, 1489, 1493, 1499,
1511, 1523, 1531, 1543, 1549, 1553, 1559, 1567, 1571, 1579, 1583, 1597,
1601, 1607, 1609, 1613, 1619, 1621, 1627, 1637, 1657, 1663, 1667, 1669, 1693, 1697, 1699,
1709, 1721, 1723, 1733, 1741, 1747, 1753, 1759, 1777, 1783, 1787, 1789,
1801, 1811, 1823, 1831, 1847, 1861, 1867, 1871, 1873, 1877, 1879, 1889,
1901, 1907, 1913, 1931, 1933, 1949, 1951, 1973, 1979, 1987, 1993, 1997, 1999 };
class BigInteger
{
typedef unsigned char byte;
public:
BigInteger(void);
BigInteger(__int64 value);
BigInteger(unsigned __int64 value);
BigInteger(const BigInteger &bi);
BigInteger(string value, int radix);
BigInteger(byte inData[], int inLen);
BigInteger(unsigned int inData[], int inLen);
BigInteger operator -();
BigInteger operator =(const BigInteger &bi2);
BigInteger operator +(BigInteger &bi2);
BigInteger operator -(BigInteger bi2);
BigInteger operator /(BigInteger bi2);
BigInteger operator *(BigInteger bi2);
void singleByteDivide(BigInteger &bi1, BigInteger &bi2, BigInteger &outQuotient, BigInteger &outRemainder);
void multiByteDivide(BigInteger &bi1, BigInteger &bi2, BigInteger &outQuotient, BigInteger &outRemainder);
BigInteger operator %(BigInteger bi2);
BigInteger operator +=(BigInteger bi2);
BigInteger operator -=(BigInteger bi2);
int bitCount();
BigInteger modPow(BigInteger exp, BigInteger n);
friend ostream& operator<<(ostream& output, BigInteger &bi1);
friend BigInteger GetPrime();
friend bool Miller_Robin(BigInteger &bi1);
friend BigInteger MultipInverse(BigInteger &bi1, BigInteger &n);
friend BigInteger extended_euclidean(BigInteger n, BigInteger m, BigInteger &x, BigInteger &y);
friend BigInteger MultipInverse(BigInteger &bi1, BigInteger &n); //求乘法逆
friend BigInteger Gcd(BigInteger &bi1, BigInteger &bi2); //求最大公约数
friend bool IsPrime (BigInteger &obj);
BigInteger BarrettReduction(BigInteger x, BigInteger n, BigInteger constant);
bool operator >=(BigInteger bi2)
{
return ((*this) == bi2 || (*this) > bi2);
}
bool operator >(BigInteger bi2);
bool operator ==(BigInteger bi2);
bool operator !=(BigInteger bi2);
int shiftRight(unsigned int buffer[], int bufLen, int shiftVal);
BigInteger operator <<(int shiftVal);
int shiftLeft(unsigned int buffer[], int bufLen, int shiftVal);
bool operator <(BigInteger bi2);
string DecToHex(unsigned int value, string format);
string ToHexString();
public:
~BigInteger(void);
public:
int dataLength;
// number of actual chars used
unsigned int *data;
};
BigInteger.cpp
#include "BigInteger.h"
BigInteger::BigInteger(void) //默认的构造函数
: dataLength(0), data(0)
{
data = new unsigned int[maxLength];
memset(data, 0, maxLength * sizeof(unsigned int));
dataLength = 1;
}
BigInteger::BigInteger(__int64 value) //用一个64位的值来初始化大数
{
data = new unsigned int[maxLength];
memset(data, 0, maxLength * sizeof(unsigned int)); //先清零
__int64 tempVal = value;
dataLength = 0;
while (value != 0 && dataLength < maxLength)
{
data[dataLength] = (unsigned int)(value & 0xFFFFFFFF); //取低位
value = value >> 32; //进位
dataLength++;
}
if (tempVal > 0) // overflow check for +ve value
{
if (value != 0 || (data[maxLength - 1] & 0x80000000) != 0)
assert(false);
}
else if (tempVal < 0) // underflow check for -ve value
{
if (value != -1 || (data[dataLength - 1] & 0x80000000) == 0)
assert(false);
}
if (dataLength == 0)
dataLength = 1;
}
BigInteger::BigInteger(unsigned __int64 value) //用一个无符号的64位整数来初始化大数
{
data = new unsigned int[maxLength];
memset(data, 0, maxLength * sizeof(unsigned int));
dataLength = 0;
while (value != 0 && dataLength < maxLength)
{
data[dataLength] = (unsigned int)(value & 0xFFFFFFFF);
value >>= 32;
dataLength++;
}
if (value != 0 || (data[maxLength - 1] & 0x80000000) != 0)
assert(false);
if (dataLength == 0) //防止输入的value=0
dataLength = 1;
}
BigInteger::BigInteger(const BigInteger &bi) //用大数初始化大数
{
data = new unsigned int[maxLength];
dataLength = bi.dataLength;
for (int i = 0; i < maxLength; i++) //考虑到有负数的情况,所以每一位都要复制
data[i] = bi.data[i];
}
BigInteger::~BigInteger(void)
{
if (data != NULL)
{
delete []data;
}
}
BigInteger::BigInteger(string value, int radix) //输入转换函数,将字符串转换成对应进制的大数
{ //一般不处理负数
BigInteger multiplier((__int64)1);
BigInteger result;
transform(value.begin(), value.end(), value.begin(), toupper); //将小写字母转换成为大写
int limit = 0;
if (value[0] == '-')
limit = 1;
for (int i = value.size() - 1; i >= limit; i--)
{
int posVal = (int)value[i];
if (posVal >= '0' && posVal <= '9') //将字符转换成数字
posVal -= '0';
else if (posVal >= 'A' && posVal <= 'Z')
posVal = (posVal - 'A') + 10;
else
posVal = 9999999; // arbitrary large 输入别的字符
if (posVal >= radix) //不能大于特定的进制,否则终止
{
assert(false);
}
else
{
result = result + (multiplier * BigInteger((__int64)posVal));
if ((i - 1) >= limit) //没有到达尾部
multiplier = multiplier * BigInteger((__int64)radix);
}
}
if (value[0] == '-') //符号最后再处理
result = -result;
if (value[0] == '-') //输入为负数,但得到的结果为正数,可能溢出了
{
if ((result.data[maxLength - 1] & 0x80000000) == 0)
assert(false);
}
else //或者说,输入为正数,得到的结果为负数,也可能溢出了
{
if ((result.data[maxLength - 1] & 0x80000000) != 0)
assert(false);
}
data = new unsigned int[maxLength];
//memset(data, 0, maxLength * sizeof(unsigned int));
for (int i = 0; i < maxLength; i++)
data[i] = result.data[i];
dataLength = result.dataLength;
}
BigInteger::BigInteger(byte inData[], int inLen) //用一个char类型的数组来初始化大数
{
dataLength = inLen >> 2; //一个unsigned int占32位,而一个unsigned char只占8位
//因此dataLength应该是至少是inLen/4,不一定整除
int leftOver = inLen & 0x3;
//取最低两位的数值,为什么要这样干呢？实际上是为了探测len是不是4的倍数,好确定dataLength的长度
if (leftOver != 0) //不能整除的话,dataLength要加1
dataLength++;
if (dataLength > maxLength)
assert(false);
data = new unsigned int[maxLength];
memset(data, 0, maxLength * sizeof(unsigned int));
for (int i = inLen - 1, j = 0; i >= 3; i -= 4, j++)
{
data[j] = (unsigned int)((inData[i - 3] << 24) + (inData[i - 2] << 16) + (inData[i - 1] << 8) + inData[i]);
//我们知道:一个unsigned int占32位,而一个unsigned char只占8位,因此四个unsigned char才能组成一个unsigned int
//因此取inData[i - 3]为前32-25位,inData[i - 2]为前24-17~~~
//i % 4 = 0 or 1 or 2 or 3 余0表示恰好表示完
}
if (leftOver == 1)
data[dataLength - 1] = (unsigned int)inData[0];
else if (leftOver == 2)
data[dataLength - 1] = (unsigned int)((inData[0] << 8) + inData[1]);
else if (leftOver == 3)
data[dataLength - 1] = (unsigned int)((inData[0] << 16) + (inData[1] << 8) + inData[2]);
while (dataLength > 1 && data[dataLength - 1] == 0)
dataLength--;
}
BigInteger::BigInteger(unsigned int inData[], int inLen) //用一个unsigned int型数组初始化大数
{
dataLength = inLen;
if (dataLength > maxLength)
assert(false);
data = new unsigned int[maxLength];
memset(data, 0, maxLength * sizeof(maxLength));
for (int i = dataLength - 1, j = 0; i >= 0; i--, j++)
data[j] = inData[i];
while (dataLength > 1 && data[dataLength - 1] == 0)
dataLength--;
}
BigInteger BigInteger::operator *(BigInteger bi2) //乘法的重载
{
BigInteger bi1(*this);
int lastPos = maxLength - 1;
bool bi1Neg = false, bi2Neg = false;
//首先对两个乘数取绝对值
try
{
if ((this->data[lastPos] & 0x80000000) != 0) //bi1为负数
{
bi1Neg = true;
bi1 = -bi1;
}
if ((bi2.data[lastPos] & 0x80000000) != 0) //bi2为负数
{
bi2Neg = true; bi2 = -bi2;
}
}
catch (...) { }
BigInteger result;
//绝对值相乘
try
{
for (int i = 0; i < bi1.dataLength; i++)
{
if (bi1.data[i] == 0) continue;
unsigned __int64 mcarry = 0;
for (int j = 0, k = i; j < bi2.dataLength; j++, k++)
{
// k = i + j
unsigned __int64 val = ((unsigned __int64)bi1.data[i] * (unsigned __int64)bi2.data[j]) + (unsigned __int64)result.data[k] + mcarry;
result.data[k] = (unsigned __int64)(val & 0xFFFFFFFF); //取低位
mcarry = (val >> 32); //进位
}
if (mcarry != 0)
result.data[i + bi2.dataLength] = (unsigned int)mcarry;
}
}
catch (...)
{
assert(false);
}
result.dataLength = bi1.dataLength + bi2.dataLength;
if (result.dataLength > maxLength)
result.dataLength = maxLength;
while (result.dataLength > 1 && result.data[result.dataLength - 1] == 0)
result.dataLength--;
// overflow check (result is -ve)溢出检查
if ((result.data[lastPos] & 0x80000000) != 0) //结果为负数
{
if (bi1Neg != bi2Neg && result.data[lastPos] == 0x80000000) //两乘数符号不同
{
// handle the special case where multiplication produces
// a max negative number in 2's complement.
if (result.dataLength == 1)
return result;
else
{
bool isMaxNeg = true;
for (int i = 0; i < result.dataLength - 1 && isMaxNeg; i++)
{
if (result.data[i] != 0)
isMaxNeg = false;
}
if (isMaxNeg)
return result;
}
}
assert(false);
}
//两乘数符号不同,结果为负数
if (bi1Neg != bi2Neg)
return -result;
return result;
}
BigInteger BigInteger::operator =(const BigInteger &bi2)
{
if (&bi2 == this)
{
return *this;
}
if (data != NULL)
{
delete []data;
data = NULL;
}
data = new unsigned int[maxLength];
memset(data, 0, maxLength * sizeof(unsigned int));
dataLength = bi2.dataLength;
for (int i = 0; i < maxLength; i++)
data[i] = bi2.data[i];
return *this;
}
BigInteger BigInteger::operator +(BigInteger &bi2)
{
int lastPos = maxLength - 1;
bool bi1Neg = false, bi2Neg = false;
BigInteger bi1(*this);
BigInteger result;
if ((this->data[lastPos] & 0x80000000) != 0) //bi1为负数
bi1Neg = true;
if ((bi2.data[lastPos] & 0x80000000) != 0) //bi2为负数
bi2Neg = true;
if(bi1Neg == false && bi2Neg == false) //bi1与bi2都是正数
{
result.dataLength = (this->dataLength > bi2.dataLength) ? this->dataLength : bi2.dataLength;
__int64 carry = 0;
for (int i = 0; i < result.dataLength; i++) //从低位开始,逐位相加
{
__int64 sum = (__int64)this->data[i] + (__int64)bi2.data[i] + carry;
carry = sum >> 32; //进位
result.data[i] = (unsigned int)(sum & 0xFFFFFFFF); //取低位结果
}
if (carry != 0 && result.dataLength < maxLength)
{
result.data[result.dataLength] = (unsigned int)(carry);
result.dataLength++;
}
while (result.dataLength > 1 && result.data[result.dataLength - 1] == 0)
result.dataLength--;
//溢出检查
if ((this->data[lastPos] & 0x80000000) == (bi2.data[lastPos] & 0x80000000) &&
(result.data[lastPos] & 0x80000000) != (this->data[lastPos] & 0x80000000))
{
assert(false);
}
return result;
}
//关键在于,负数全部要转化成为正数来做
if(bi1Neg == false && bi2Neg == true) //bi1正,bi2负
{
BigInteger bi3 = -bi2;
if(bi1 > bi3)
{
result = bi1 - bi3;
return result;
}
else
{
result = -(bi3 - bi1);
return result;
}
}
if(bi1Neg == true && bi2Neg == false) //bi1负,bi2正
{
BigInteger bi3 = -bi1;
if(bi3 > bi2)
{
result = -(bi3 - bi2);
return result;
}
else
{
result = bi2 - bi3;
return result;
}
}
if(bi1Neg == true && bi2Neg == true) //bi1负,bi2负
{
result = - ((-bi1) + (-bi2));
return result;
}
}
BigInteger BigInteger::operator -()
{
//逐位取反并+1
if (this->dataLength == 1 && this->data[0] == 0)
return *this;
BigInteger result(*this);
for (int i = 0; i < maxLength; i++)
result.data[i] = (unsigned int)(~(this->data[i])); //取反
__int64 val, carry = 1;
int index = 0;
while (carry != 0 && index < maxLength) //+1；
{
val = (__int64)(result.data[index]);
val++; //由于值加了1个1,往前面的进位最多也只是1个1,因此val++就完了
result.data[index] = (unsigned int)(val & 0xFFFFFFFF); //取低位部分
carry = val >> 32; //进位
index++;
}
if ((this->data[maxLength - 1] & 0x80000000) == (result.data[maxLength - 1] & 0x80000000))
result.dataLength = maxLength;
while (result.dataLength > 1 && result.data[result.dataLength - 1] == 0)
result.dataLength--;
return result;
}
BigInteger BigInteger::modPow(BigInteger exp, BigInteger n) //求this^exp mod n
{
if ((exp.data[maxLength - 1] & 0x80000000) != 0) //指数是负数
return BigInteger((__int64)0);
BigInteger resultNum((__int64)1);
BigInteger tempNum;
bool thisNegative = false;
if ((this->data[maxLength - 1] & 0x80000000) != 0) //底数是负数
{
tempNum = -(*this) % n;
thisNegative = true;
}
else
tempNum = (*this) % n; //保证(tempNum * tempNum) < b^(2k)
if ((n.data[maxLength - 1] & 0x80000000) != 0) //n为负
n = -n;
//计算 constant = b^(2k) / m
//constant主要用于后面的Baeert Reduction算法
BigInteger constant;
int i = n.dataLength << 1;
constant.data[i] = 0x00000001;
constant.dataLength = i + 1;
constant = constant / n;
int totalBits = exp.bitCount();
int count = 0;
//平方乘法算法
for (int pos = 0; pos < exp.dataLength; pos++)
{
unsigned int mask = 0x01;
for (int index = 0; index < 32; index++)
{
if ((exp.data[pos] & mask) != 0) //某一个bit不为0
resultNum = BarrettReduction(resultNum * tempNum, n, constant);
//resultNum = resultNum * tempNum mod n
mask <<= 1; //不断左移
tempNum = BarrettReduction(tempNum * tempNum, n, constant);
//tempNum = tempNum * tempNum mod n
if (tempNum.dataLength == 1 && tempNum.data[0] == 1)
{
if (thisNegative && (exp.data[0] & 0x1) != 0) //指数为奇数
return -resultNum;
return resultNum;
}
count++;
if (count == totalBits)
break;
}
}
if (thisNegative && (exp.data[0] & 0x1) != 0) //底数为负数,指数为奇数
return -resultNum;
return resultNum;
}
int BigInteger::bitCount() //计算字节数
{
while (dataLength > 1 && data[dataLength - 1] == 0)
dataLength--;
unsigned int value = data[dataLength - 1];
unsigned int mask = 0x80000000;
int bits = 32;
while (bits > 0 && (value & mask) == 0) //计算最高位的bit
{
bits--;
mask >>= 1;
}
bits += ((dataLength - 1) << 5); //余下的位都有32bit
//左移5位,相当于乘以32,即2^5
return bits;
}
BigInteger BigInteger::BarrettReduction(BigInteger x, BigInteger n, BigInteger constant)
{
//算法,Baeert Reduction算法,在计算大规模的除法运算时很有优势
//原理如下
//Z mod N=Z-[Z/N]*N=Z-{[Z/b^(n-1)]*[b^2n/N]/b^(n+1)}*N=Z-q*N
//q=[Z/b^(n-1)]*[b^2n/N]/b^(n+1)
//其中,[]表示取整运算,A^B表示A的B次幂
int k = n.dataLength,
kPlusOne = k + 1,
kMinusOne = k - 1;
BigInteger q1;
// q1 = x / b^(k-1)
for (int i = kMinusOne, j = 0; i < x.dataLength; i++, j++)
q1.data[j] = x.data[i];
q1.dataLength = x.dataLength - kMinusOne;
if (q1.dataLength <= 0)
q1.dataLength = 1;
BigInteger q2 = q1 * constant;
BigInteger q3;
// q3 = q2 / b^(k+1)
for (int i = kPlusOne, j = 0; i < q2.dataLength; i++, j++)
q3.data[j] = q2.data[i];
q3.dataLength = q2.dataLength - kPlusOne;
if (q3.dataLength <= 0)
q3.dataLength = 1;
// r1 = x mod b^(k+1)
// i.e. keep the lowest (k+1) words
BigInteger r1;
int lengthToCopy = (x.dataLength > kPlusOne) ? kPlusOne : x.dataLength;
for (int i = 0; i < lengthToCopy; i++)
r1.data[i] = x.data[i];
r1.dataLength = lengthToCopy;
// r2 = (q3 * n) mod b^(k+1)
// partial multiplication of q3 and n
BigInteger r2;
for (int i = 0; i < q3.dataLength; i++)
{
if (q3.data[i] == 0) continue;
unsigned __int64 mcarry = 0;
int t = i;
for (int j = 0; j < n.dataLength && t < kPlusOne; j++, t++)
{
// t = i + j
unsigned __int64 val = ((unsigned __int64)q3.data[i] * (unsigned __int64)n.data[j]) +
(unsigned __int64)r2.data[t] + mcarry;
r2.data[t] = (unsigned int)(val & 0xFFFFFFFF);
mcarry = (val >> 32);
}
if (t < kPlusOne)
r2.data[t] = (unsigned int)mcarry;
}
r2.dataLength = kPlusOne;
while (r2.dataLength > 1 && r2.data[r2.dataLength - 1] == 0)
r2.dataLength--;
r1 -= r2;
if ((r1.data[maxLength - 1] & 0x80000000) != 0) // negative
{
BigInteger val;
val.data[kPlusOne] = 0x00000001;
val.dataLength = kPlusOne + 1;
r1 += val;
}
while (r1 >= n)
r1 -= n;
return r1;
}
bool BigInteger::operator >(BigInteger bi2)
{
int pos = maxLength - 1;
BigInteger bi1(*this);
// bi1 is negative, bi2 is positive
if ((bi1.data[pos] & 0x80000000) != 0 && (bi2.data[pos] & 0x80000000) == 0)
return false;
// bi1 is positive, bi2 is negative
else if ((bi1.data[pos] & 0x80000000) == 0 && (bi2.data[pos] & 0x80000000) != 0)
return true;
// same sign
int len = (bi1.dataLength > bi2.dataLength) ? bi1.dataLength : bi2.dataLength;
for (pos = len - 1; pos >= 0 && bi1.data[pos] == bi2.data[pos]; pos--) ;
if (pos >= 0)
{
if (bi1.data[pos] > bi2.data[pos])
return true;
return false;
}
return false;
}
bool BigInteger::operator ==(BigInteger bi2)
{
if (this->dataLength != bi2.dataLength)
return false;
for (int i = 0; i < this->dataLength; i++)
{
if (this->data[i] != bi2.data[i])
return false;
}
return true;
}
bool BigInteger::operator !=(BigInteger bi2)
{
if(this->dataLength != bi2.dataLength)
return true;
for(int i = 0; i < this->dataLength; i++)
{
if(this->data[i] != bi2.data[i])
return true;
}
return false;
}
BigInteger BigInteger::operator %(BigInteger bi2)
{
BigInteger bi1(*this);
BigInteger quotient;
BigInteger remainder(bi1);
int lastPos = maxLength - 1;
bool dividendNeg = false;
if ((bi1.data[lastPos] & 0x80000000) != 0) // bi1 negative
{
bi1 = -bi1;
dividendNeg = true;
}
if ((bi2.data[lastPos] & 0x80000000) != 0) // bi2 negative
bi2 = -bi2;
if (bi1 < bi2)
{
return remainder;
}
else
{
if (bi2.dataLength == 1)
singleByteDivide(bi1, bi2, quotient, remainder); //bi2只占一个位置时,用singleByteDivide更快
else
multiByteDivide(bi1, bi2, quotient, remainder); //bi2占多个位置时,用multiByteDivide更快
if (dividendNeg)
return -remainder;
return remainder;
}
}
void BigInteger::singleByteDivide(BigInteger &bi1, BigInteger &bi2,
BigInteger &outQuotient, BigInteger &outRemainder)
{//outQuotient商,outRemainder余数
unsigned int result[maxLength]; //用来存储结果
memset(result, 0, sizeof(unsigned int) * maxLength);
int resultPos = 0;
for (int i = 0; i < maxLength; i++) //将bi1复制至outRemainder
outRemainder.data[i] = bi1.data[i];
outRemainder.dataLength = bi1.dataLength;
while (outRemainder.dataLength > 1 && outRemainder.data[outRemainder.dataLength - 1] == 0)
outRemainder.dataLength--;
unsigned __int64 divisor = (unsigned __int64)bi2.data[0];
int pos = outRemainder.dataLength - 1;
unsigned __int64 dividend = (unsigned __int64)outRemainder.data[pos]; //取最高位的数值
if (dividend >= divisor) //被除数>除数
{
unsigned __int64 quotient = dividend / divisor;
result[resultPos++] = (unsigned __int64)quotient; //结果
outRemainder.data[pos] = (unsigned __int64)(dividend % divisor); //余数
}
pos--;
while (pos >= 0)
{
dividend = ((unsigned __int64)outRemainder.data[pos + 1] << 32) + (unsigned __int64)outRemainder.data[pos]; //前一位的余数和这一位的值相加
unsigned __int64 quotient = dividend / divisor; //得到结果
result[resultPos++] = (unsigned int)quotient; //结果取低位
outRemainder.data[pos + 1] = 0; //前一位的余数清零
outRemainder.data[pos--] = (unsigned int)(dividend % divisor); //得到这一位的余数
}
outQuotient.dataLength = resultPos; //商的长度是resultPos的长度
int j = 0;
for (int i = outQuotient.dataLength - 1; i >= 0; i--, j++) //将商反转过来
outQuotient.data[j] = result[i];
for (; j < maxLength; j++) //商的其余位都要置0
outQuotient.data[j] = 0;
while (outQuotient.dataLength > 1 && outQuotient.data[outQuotient.dataLength - 1] == 0)
outQuotient.dataLength--;
if (outQuotient.dataLength == 0)
outQuotient.dataLength = 1;
while (outRemainder.dataLength > 1 && outRemainder.data[outRemainder.dataLength - 1] == 0)
outRemainder.dataLength--;
}
void BigInteger::multiByteDivide(BigInteger &bi1, BigInteger &bi2,
BigInteger &outQuotient, BigInteger &outRemainder)
{
unsigned int result[maxLength];
memset(result, 0, sizeof(unsigned int) * maxLength); //结果置零
int remainderLen = bi1.dataLength + 1; //余数长度
unsigned int *remainder = new unsigned int[remainderLen];
memset(remainder, 0, sizeof(unsigned int) * remainderLen); //余数置零
unsigned int mask = 0x80000000;
unsigned int val = bi2.data[bi2.dataLength - 1];
int shift = 0, resultPos = 0;
while (mask != 0 && (val & mask) == 0)
{
shift++; mask >>= 1;
}
//最高位从高到低找出shift个0位
for (int i = 0; i < bi1.dataLength; i++)
remainder[i] = bi1.data[i]; //将bi1复制到remainder之中
this->shiftLeft(remainder, remainderLen, shift); //remainder左移shift位
bi2 = bi2 << shift; //向左移shift位,将空位填满
//由于两个数都扩大了相同的倍数,所以结果不变
int j = remainderLen - bi2.dataLength; //j表示两个数长度的差值,也是要计算的次数
int pos = remainderLen - 1; //pos指示余数的最高位的位置,现在pos=bi1.dataLength
//以下的步骤并没有别的意思,主要是用来试商
unsigned __int64 firstDivisorByte = bi2.data[bi2.dataLength - 1]; //第一个除数
unsigned __int64 secondDivisorByte = bi2.data[bi2.dataLength - 2]; //第二个除数
int divisorLen = bi2.dataLength + 1; //除数的长度
unsigned int *dividendPart = new unsigned int[divisorLen]; //起名为除数的部分
memset(dividendPart, 0, sizeof(unsigned int) * divisorLen);
while (j > 0)
{
unsigned __int64 dividend = ((unsigned __int64)remainder[pos] << 32) + (unsigned __int64)remainder[pos - 1]; //取余数的高两位
unsigned __int64 q_hat = dividend / firstDivisorByte; //得到一个商
unsigned __int64 r_hat = dividend % firstDivisorByte; //以及一个余数
bool done = false; //表示没有做完
while (!done)
{
done = true;
if (q_hat == 0x100000000 || (q_hat * secondDivisorByte) > ((r_hat << 32) + remainder[pos - 2])) //这里主要用来调整商的大小
//(q_hat * secondDivisorByte) > ((r_hat << 32) + remainder[pos - 2]))是害怕上的商过大,减之后变为负数
//商q_hat也不能超过32bit
{
q_hat--; //否则的话,就商小一点,余数大一点
r_hat += firstDivisorByte;
if (r_hat < 0x100000000) //如果余数小于32bit,就继续循环
done = false;
}
}
for (int h = 0; h < divisorLen; h++) //取被除数的高位部分,高位部分长度与除数长度一致
dividendPart[h] = remainder[pos - h];
BigInteger kk(dividendPart, divisorLen);
BigInteger ss = bi2 * BigInteger((__int64)q_hat);
while (ss > kk) //调节商的大小
{
q_hat--;
ss -= bi2;
}
BigInteger yy = kk - ss; //得到余数
for (int h = 0; h < divisorLen; h++) //将yy高位和remainder低位拼接起来,得到余数
remainder[pos - h] = yy.data[bi2.dataLength - h]; //取得真正的余数
result[resultPos++] = (unsigned int)q_hat;
pos--;
j--;
}
outQuotient.dataLength = resultPos;
int y = 0;
for (int x = outQuotient.dataLength - 1; x >= 0; x--, y++) //将商反转过来
outQuotient.data[y] = result[x];
for (; y < maxLength; y++) //商的其余位都要置0
outQuotient.data[y] = 0;
while (outQuotient.dataLength > 1 && outQuotient.data[outQuotient.dataLength - 1] == 0)
outQuotient.dataLength--;
if (outQuotient.dataLength == 0)
outQuotient.dataLength = 1;
outRemainder.dataLength = this->shiftRight(remainder, remainderLen, shift);
for (y = 0; y < outRemainder.dataLength; y++)
outRemainder.data[y] = remainder[y];
for (; y < maxLength; y++)
outRemainder.data[y] = 0;
delete []remainder;
delete []dividendPart;
}
int BigInteger::shiftRight(unsigned int buffer[], int bufferLen,int shiftVal) //右移操作
{//自己用图画模拟一下移位操作,就能很快明白意义了
int shiftAmount = 32;
int invShift = 0;
int bufLen = bufferLen;
while (bufLen > 1 && buffer[bufLen - 1] == 0)
bufLen--;
for (int count = shiftVal; count > 0; )
{
if (count < shiftAmount)
{
shiftAmount = count;
invShift = 32 - shiftAmount;
}
unsigned __int64 carry = 0;
for (int i = bufLen - 1; i >= 0; i--)
{
unsigned __int64 val = ((unsigned __int64)buffer[i]) >> shiftAmount;
val |= carry;
carry = ((unsigned __int64)buffer[i]) << invShift;
buffer[i] = (unsigned int)(val);
}
count -= shiftAmount;
}
while (bufLen > 1 && buffer[bufLen - 1] == 0)
bufLen--;
return bufLen;
}
BigInteger BigInteger::operator <<(int shiftVal)
{
BigInteger result(*this);
result.dataLength = shiftLeft(result.data, maxLength, shiftVal);
return result;
}
int BigInteger::shiftLeft(unsigned int buffer[], int bufferLen, int shiftVal)
{
int shiftAmount = 32;
int bufLen = bufferLen;
while (bufLen > 1 && buffer[bufLen - 1] == 0)
bufLen--;
for (int count = shiftVal; count > 0; )
{
if (count < shiftAmount)
shiftAmount = count;
unsigned __int64 carry = 0;
for (int i = 0; i < bufLen; i++)
{
unsigned __int64 val = ((unsigned __int64)buffer[i]) << shiftAmount;
val |= carry;
buffer[i] = (unsigned int)(val & 0xFFFFFFFF);
carry = val >> 32;
}
if (carry != 0)
{
if (bufLen + 1 <= bufferLen)
{
buffer[bufLen] = (unsigned int)carry;
bufLen++;
}
}
count -= shiftAmount;
}
return bufLen;
}
bool BigInteger::operator <(BigInteger bi2)
{
BigInteger bi1(*this);
int pos = maxLength - 1;
// bi1 is negative, bi2 is positive
if ((bi1.data[pos] & 0x80000000) != 0 && (bi2.data[pos] & 0x80000000) == 0)
return true;
// bi1 is positive, bi2 is negative
else if ((bi1.data[pos] & 0x80000000) == 0 && (bi2.data[pos] & 0x80000000) != 0)
return false;
// same sign
int len = (bi1.dataLength > bi2.dataLength) ? bi1.dataLength : bi2.dataLength;
for (pos = len - 1; pos >= 0 && bi1.data[pos] == bi2.data[pos]; pos--) ;
if (pos >= 0)
{
if (bi1.data[pos] < bi2.data[pos])
return true;
return false;
}
return false;
}
BigInteger BigInteger::operator +=(BigInteger bi2)
{
*this = *this + bi2;
return *this;
}
BigInteger BigInteger::operator /(BigInteger bi2)
{
BigInteger bi1(*this);
BigInteger quotient;
BigInteger remainder;
int lastPos = maxLength - 1;
bool divisorNeg = false, dividendNeg = false;
if ((bi1.data[lastPos] & 0x80000000) != 0) // bi1 negative
{
bi1 = -bi1;
dividendNeg = true;
}
if ((bi2.data[lastPos] & 0x80000000) != 0) // bi2 negative
{
bi2 = -bi2;
divisorNeg = true;
}
if (bi1 < bi2)
{
return quotient;
}
else
{
if (bi2.dataLength == 1)
singleByteDivide(bi1, bi2, quotient, remainder);
else
multiByteDivide(bi1, bi2, quotient, remainder);
if (dividendNeg != divisorNeg)
return -quotient;
return quotient;
}
}
BigInteger BigInteger::operator -=(BigInteger bi2)
{
*this = *this - bi2;
return *this;
}
BigInteger BigInteger::operator -(BigInteger bi2) //减法的重载
{
BigInteger bi1(*this);
BigInteger result;
int lastPos = maxLength - 1;
bool bi1Neg = false, bi2Neg = false;
if ((this->data[lastPos] & 0x80000000) != 0) // bi1 negative
bi1Neg = true;
if ((bi2.data[lastPos] & 0x80000000) != 0) // bi1 negative
bi2Neg = true;
if(bi1Neg == false && bi2Neg == false) //bi1,bi2都为正数
{
if(bi1 < bi2)
{
result = -(bi2 - bi1);
return result;
}
result.dataLength = (bi1.dataLength > bi2.dataLength) ? bi1.dataLength : bi2.dataLength;
__int64 carryIn = 0;
for (int i = 0; i < result.dataLength; i++) //从低位开始减
{
__int64 diff;
diff = (__int64)bi1.data[i] - (__int64)bi2.data[i] - carryIn;
result.data[i] = (unsigned int)(diff & 0xFFFFFFFF);
if (diff < 0)
carryIn = 1;
else
carryIn = 0;
}
if (carryIn != 0)
{
for (int i = result.dataLength; i < maxLength; i++)
result.data[i] = 0xFFFFFFFF;
result.dataLength = maxLength;
}
// fixed in v1.03 to give correct datalength for a - (-b)
while (result.dataLength > 1 && result.data[result.dataLength - 1] == 0)
result.dataLength--;
// overflow check
if ((bi1.data[lastPos] & 0x80000000) != (bi2.data[lastPos] & 0x80000000) &&
(result.data[lastPos] & 0x80000000) != (bi1.data[lastPos] & 0x80000000))
{
assert(false);
}
return result;
}
if(bi1Neg == true && bi2Neg == false) //bi1负,bi2正
{
result = -(-bi1 + bi2);
return result;
}
if(bi1Neg == false && bi2Neg == true) //bi1正,bi2负
{
result = bi1 + (-bi2);
return result;
}
if(bi1Neg == true && bi2Neg == true) //bi1,bi2皆为负
{
BigInteger bi3 = -bi1, bi4 = -bi2;
if(bi3 > bi4)
{
result = -(bi3 - bi4);
return result;
}
else
{
result = bi4 - bi3;
return result;
}
}
}
string BigInteger::DecToHex(unsigned int value, string format) //10进制数转换成16进制数,并用string表示
{
string HexStr;
int a[100];
int i = 0;
int m = 0;
int mod = 0;
char hex[16]={'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'};
while(value > 0)
{
mod = value % 16;
a[i++] = mod;
value = value/16;
}
for(i = i - 1; i >= 0; i--)
{
m=a[i];
HexStr.push_back(hex[m]);
}
while (format == string("X8") && HexStr.size() < 8)
{
HexStr = "0" + HexStr;
}
return HexStr;
}
string BigInteger::ToHexString() //功能：将一个大数用16进制的string表示出来
{
string result = DecToHex(data[dataLength - 1], string("X"));
for (int i = dataLength - 2; i >= 0; i--)
{
result += DecToHex(data[i], string("X8"));
}
return result;
}
ostream& operator<<(ostream& output, BigInteger &obj)//以16进制输出数值
{
//if ((obj.data[obj.dataLength-1] & 0x80000000) != 0) // bi1 negative
for(int i = obj.dataLength - 1; i >= 0; i--)
output << hex << obj.data[i];
return output;
}
bool Miller_Robin(BigInteger &bi1) //Miller_Robin算法
{
BigInteger one((__int64)1), two((__int64)2), sum, a, b, temp;
int k = 0, len = primeLength / 2;
temp = sum = bi1 - one;
while((sum.data[0] & 0x00000001) == 0) //只要sum不为奇数,sum就一直往右移
{
sum.dataLength = sum.shiftRight(sum.data, maxLength, 1); //右移一位
k++;
}
//sum即为要求的奇数,k即是要求的2的次数
srand((unsigned)time(0));
for(int i = 0; i < len; i++)
{
a.data[i] =(unsigned int)rand ();
if(a.data[i] != 0) a.dataLength = i + 1;
}
b = a.modPow(sum, bi1); //b = a^m mod bi1
if (b == one) return true;
for(int i = 0; i < k; i++)
{
if(b == temp) return true;
else b = b.modPow(two, bi1); //b = b^2 mod bi1
}
return false;
}
bool IsPrime (BigInteger &obj)
{
BigInteger zero;
for(int i = 0; i < 303; i++) //先用一些素数对这个整数进行筛选
{
BigInteger prime((__int64)primesBelow2000[i]);
if(obj % prime == zero)
return false;
}
cout << "第一轮素性检验通过… … … …" << endl;
cout << "正在进行Miller_Robin素性检验… … … …" << endl;
if(Miller_Robin(obj)) //进行1次Miller_Robin检验
return true; //通过了就返回result
return false;//表明result是合数,没有通过检验
}
BigInteger GetPrime()
{
BigInteger one((__int64)1), two((__int64)2), result;
srand((unsigned)time(0));
//随机产生一个大整数
for(int i = 0; i < primeLength; i++)
{
result.data[i] =(unsigned int)rand();
if(result.data[i] != 0)
result.dataLength = i + 1;
}
result.data[0] |= 0x00000001; //保证这个整数为奇数
while(!IsPrime(result)) //如果没有通过检验,就+2,继续检验
{
result = result + two;
cout << "检验没有通过,进行下一个数的检验,运行之中… … … …" << endl;
cout << endl;
}
return result;
}
BigInteger extended_euclidean(BigInteger n, BigInteger m, BigInteger &x, BigInteger &y) //扩展的欧几里德算法
{
BigInteger x1((__int64)1), x2, x3(n);
BigInteger y1, y2((__int64)1), y3(m);
BigInteger zero;
while(x3 % y3 != zero)
{
BigInteger d = x3 / y3;
BigInteger t1, t2, t3;
t1 = x1 - d * y1;
t2 = x2 - d * y2;
t3 = x3 - d * y3;
x1 = y1; x2 = y2; x3 = y3;
y1 = t1; y2 = t2; y3 = t3;
}
x = y1; y = y2;
return y3;
}
/*
BigInteger extended_euclidean(BigInteger n,BigInteger m,BigInteger &x,BigInteger &y)
{
BigInteger zero, one((__int64)1);
if(m == zero) { x = one; y = zero; return n; }
BigInteger g = extended_euclidean(m, n%m, x, y);
BigInteger t = x - n / m * y;
x = y;
y = t;
return g;
}
*/
BigInteger Gcd(BigInteger &bi1, BigInteger &bi2)
{
BigInteger x, y;
BigInteger g = extended_euclidean(bi1, bi2, x, y);
return g;
}
BigInteger MultipInverse(BigInteger &bi1, BigInteger &n) //求乘法逆元
{
BigInteger x, y;
extended_euclidean(bi1, n, x, y);
if ((x.data[maxLength-1] & 0x80000000) != 0) // x negative
x = x + n;
// unsigned int i = x.data[maxLength-1] & 0x80000000;
// cout << i << endl;
return x;
}
还需要一个用于计算消息hash值的MD5~
md5.h
#include <stdio.h>
// #include <stdint.h>
#include <string.h>
#include <assert.h>
#define ROTL32(dword, n) ((dword) << (n) ^ ((dword) >> (32 - (n))))
/*MD5的结果数据长度*/
static const unsigned int MD5_HASH_SIZE = 16;
/*每次处理的BLOCK的大小*/
static const unsigned int MD5_BLOCK_SIZE = 64;
//================================================================================================
/*MD5的算法*/
/*md5算法的上下文，保存一些状态，中间数据，结果*/
typedef struct md5_ctx
{
/*处理的数据的长度*/
unsigned __int64 length;
/*还没有处理的数据长度*/
unsigned __int64 unprocessed;
/*取得的HASH结果（中间数据）*/
unsigned int hash[4];
} md5_ctx;
static void md5_init(md5_ctx *ctx)
{
ctx->length = 0;
ctx->unprocessed = 0;
/* initialize state */
/*不要奇怪为什么初始数值与参考数值不同,这是因为我们使用的数据结构的关系,大的在低位，小的在高位,8位8位一读*/
ctx->hash[0] = 0x67452301; /*应该这样读0x01234567*/
ctx->hash[1] = 0xefcdab89; /*0x89abcdef*/
ctx->hash[2] = 0x98badcfe; /*0xfedcba98*/
ctx->hash[3] = 0x10325476; /*0x76543210*/
}
#define MD5_F(x, y, z) ((((y) ^ (z)) & (x)) ^ (z))
#define MD5_G(x, y, z) (((x) & (z)) | ((y) & (~z)))
#define MD5_H(x, y, z) ((x) ^ (y) ^ (z))
#define MD5_I(x, y, z) ((y) ^ ((x) | (~z)))
/* 一共4轮，每一轮使用不同函数*/
#define MD5_ROUND1(a, b, c, d, x, s, ac) { \
(a) += MD5_F((b), (c), (d)) + (x) + (ac); \
(a) = ROTL32((a), (s)); \
(a) += (b); \
}
#define MD5_ROUND2(a, b, c, d, x, s, ac) { \
(a) += MD5_G((b), (c), (d)) + (x) + (ac); \
(a) = ROTL32((a), (s)); \
(a) += (b); \
}
#define MD5_ROUND3(a, b, c, d, x, s, ac) { \
(a) += MD5_H((b), (c), (d)) + (x) + (ac); \
(a) = ROTL32((a), (s)); \
(a) += (b); \
}
#define MD5_ROUND4(a, b, c, d, x, s, ac) { \
(a) += MD5_I((b), (c), (d)) + (x) + (ac); \
(a) = ROTL32((a), (s)); \
(a) += (b); \
}
static void md5_process_block(unsigned int state[4], const unsigned int block[MD5_BLOCK_SIZE / 4])
{
register unsigned a, b, c, d;
a = state[0];
b = state[1];
c = state[2];
d = state[3];
const unsigned int *x = block;
MD5_ROUND1(a, b, c, d, x[ 0], 7, 0xd76aa478);
MD5_ROUND1(d, a, b, c, x[ 1], 12, 0xe8c7b756);
MD5_ROUND1(c, d, a, b, x[ 2], 17, 0x242070db);
MD5_ROUND1(b, c, d, a, x[ 3], 22, 0xc1bdceee);
MD5_ROUND1(a, b, c, d, x[ 4], 7, 0xf57c0faf);
MD5_ROUND1(d, a, b, c, x[ 5], 12, 0x4787c62a);
MD5_ROUND1(c, d, a, b, x[ 6], 17, 0xa8304613);
MD5_ROUND1(b, c, d, a, x[ 7], 22, 0xfd469501);
MD5_ROUND1(a, b, c, d, x[ 8], 7, 0x698098d8);
MD5_ROUND1(d, a, b, c, x[ 9], 12, 0x8b44f7af);
MD5_ROUND1(c, d, a, b, x[10], 17, 0xffff5bb1);
MD5_ROUND1(b, c, d, a, x[11], 22, 0x895cd7be);
MD5_ROUND1(a, b, c, d, x[12], 7, 0x6b901122);
MD5_ROUND1(d, a, b, c, x[13], 12, 0xfd987193);
MD5_ROUND1(c, d, a, b, x[14], 17, 0xa679438e);
MD5_ROUND1(b, c, d, a, x[15], 22, 0x49b40821);
MD5_ROUND2(a, b, c, d, x[ 1], 5, 0xf61e2562);
MD5_ROUND2(d, a, b, c, x[ 6], 9, 0xc040b340);
MD5_ROUND2(c, d, a, b, x[11], 14, 0x265e5a51);
MD5_ROUND2(b, c, d, a, x[ 0], 20, 0xe9b6c7aa);
MD5_ROUND2(a, b, c, d, x[ 5], 5, 0xd62f105d);
MD5_ROUND2(d, a, b, c, x[10], 9, 0x2441453);
MD5_ROUND2(c, d, a, b, x[15], 14, 0xd8a1e681);
MD5_ROUND2(b, c, d, a, x[ 4], 20, 0xe7d3fbc8);
MD5_ROUND2(a, b, c, d, x[ 9], 5, 0x21e1cde6);
MD5_ROUND2(d, a, b, c, x[14], 9, 0xc33707d6);
MD5_ROUND2(c, d, a, b, x[ 3], 14, 0xf4d50d87);
MD5_ROUND2(b, c, d, a, x[ 8], 20, 0x455a14ed);
MD5_ROUND2(a, b, c, d, x[13], 5, 0xa9e3e905);
MD5_ROUND2(d, a, b, c, x[ 2], 9, 0xfcefa3f8);
MD5_ROUND2(c, d, a, b, x[ 7], 14, 0x676f02d9);
MD5_ROUND2(b, c, d, a, x[12], 20, 0x8d2a4c8a);
MD5_ROUND3(a, b, c, d, x[ 5], 4, 0xfffa3942);
MD5_ROUND3(d, a, b, c, x[ 8], 11, 0x8771f681);
MD5_ROUND3(c, d, a, b, x[11], 16, 0x6d9d6122);
MD5_ROUND3(b, c, d, a, x[14], 23, 0xfde5380c);
MD5_ROUND3(a, b, c, d, x[ 1], 4, 0xa4beea44);
MD5_ROUND3(d, a, b, c, x[ 4], 11, 0x4bdecfa9);
MD5_ROUND3(c, d, a, b, x[ 7], 16, 0xf6bb4b60);
MD5_ROUND3(b, c, d, a, x[10], 23, 0xbebfbc70);
MD5_ROUND3(a, b, c, d, x[13], 4, 0x289b7ec6);
MD5_ROUND3(d, a, b, c, x[ 0], 11, 0xeaa127fa);
MD5_ROUND3(c, d, a, b, x[ 3], 16, 0xd4ef3085);
MD5_ROUND3(b, c, d, a, x[ 6], 23, 0x4881d05);
MD5_ROUND3(a, b, c, d, x[ 9], 4, 0xd9d4d039);
MD5_ROUND3(d, a, b, c, x[12], 11, 0xe6db99e5);
MD5_ROUND3(c, d, a, b, x[15], 16, 0x1fa27cf8);
MD5_ROUND3(b, c, d, a, x[ 2], 23, 0xc4ac5665);
MD5_ROUND4(a, b, c, d, x[ 0], 6, 0xf4292244);
MD5_ROUND4(d, a, b, c, x[ 7], 10, 0x432aff97);
MD5_ROUND4(c, d, a, b, x[14], 15, 0xab9423a7);
MD5_ROUND4(b, c, d, a, x[ 5], 21, 0xfc93a039);
MD5_ROUND4(a, b, c, d, x[12], 6, 0x655b59c3);
MD5_ROUND4(d, a, b, c, x[ 3], 10, 0x8f0ccc92);
MD5_ROUND4(c, d, a, b, x[10], 15, 0xffeff47d);
MD5_ROUND4(b, c, d, a, x[ 1], 21, 0x85845dd1);
MD5_ROUND4(a, b, c, d, x[ 8], 6, 0x6fa87e4f);
MD5_ROUND4(d, a, b, c, x[15], 10, 0xfe2ce6e0);
MD5_ROUND4(c, d, a, b, x[ 6], 15, 0xa3014314);
MD5_ROUND4(b, c, d, a, x[13], 21, 0x4e0811a1);
MD5_ROUND4(a, b, c, d, x[ 4], 6, 0xf7537e82);
MD5_ROUND4(d, a, b, c, x[11], 10, 0xbd3af235);
MD5_ROUND4(c, d, a, b, x[ 2], 15, 0x2ad7d2bb);
MD5_ROUND4(b, c, d, a, x[ 9], 21, 0xeb86d391);
state[0] += a;
state[1] += b;
state[2] += c;
state[3] += d;
}
static void md5_update(md5_ctx *ctx, const unsigned char *buf, unsigned int size)
{
/*为什么不是=，因为在某些环境下，可以多次调用zen_md5_update，但这种情况，必须保证前面的调用，每次都没有unprocessed*/
ctx->length += size;
/*每个处理的块都是64字节*/
while (size >= MD5_BLOCK_SIZE)
{
md5_process_block(ctx->hash, reinterpret_cast<const unsigned int *>(buf));
buf += MD5_BLOCK_SIZE; /*buf指针每一次向后挪动64*/
size -= MD5_BLOCK_SIZE; /*每一次处理64个字符*/
}
ctx->unprocessed = size; /*未处理的字符数数目记录下来*/
}
static void md5_final(md5_ctx *ctx, const unsigned char *buf, unsigned int size, unsigned char *result)
{
unsigned int message[MD5_BLOCK_SIZE / 4];
memset(message, 0 ,(MD5_BLOCK_SIZE / 4) * sizeof(unsigned int));
/*保存剩余的数据，我们要拼出最后1个（或者两个）要处理的块，前面的算法保证了，最后一个块肯定小于64个字节*/
if (ctx->unprocessed)
{
memcpy(message, buf + size - ctx->unprocessed, static_cast<unsigned int>( ctx->unprocessed));
/*================================================================================
这里的memcpy复制很有趣,是按照字节复制比如说buf --- 0x11 0x14 0xab 0x23 0xcd |
ctx>unprocessed_=5 现在copy至 message --- 0x23ab1411 0x000000cd
这样的话,下面的也很好解释了!
=================================================================================*/
}
/*=================================================================================
用法：static_cast < type-id > ( expression )
该运算符把expression转换为type-id类型
==================================================================================*/
/*得到0x80要添加在的位置（在unsigned int 数组中）*/
unsigned int index = ((unsigned int)ctx->length & 63) >> 2;
/*一次性处理64个unsigned int型数据,(unsigned int)ctx->length_ & 63求出余下多少未处理的字符*/
unsigned int shift = ((unsigned int)ctx->length & 3) * 8;
/*一个message里面可以放置4个字符数据,找到应该移动的位数*/
/*添加0x80进去，并且把余下的空间补充0*/
message[index++] ^= 0x80 << shift; /*^ 位异或*/
/*如果这个block还无法处理，其后面的长度无法容纳长度64bit，那么先处理这个block*/
if (index > 14)
{
while (index < 16)
{
message[index++] = 0;
}
md5_process_block(ctx->hash, message);
index = 0;
}
/*补0*/
while (index < 14)
{
message[index++] = 0;
}
/*保存长度，注意是bit位的长度*/
unsigned __int64 data_len = (ctx->length) << 3;
message[14] = (unsigned int) (data_len & 0x00000000FFFFFFFF);
message[15] = (unsigned int) ((data_len & 0xFFFFFFFF00000000ULL) >> 32);
md5_process_block(ctx->hash, message);
memcpy(result, &ctx->hash, MD5_HASH_SIZE);
}
unsigned char* md5(const unsigned char *buf, unsigned int size, unsigned char result[MD5_HASH_SIZE])
{
md5_ctx ctx;
md5_init(&ctx); /*初始化*/
md5_update(&ctx, buf, size);
md5_final(&ctx, buf, size, result);
return result;
}
然后才是ELGamal的实现~
ELGamal.h
#include <string>
#include "BigInteger.h"
#include "md5.h"
/*
*事先说明一句：由于大素数的本原元很难求得，所以这里的数字签名所需要的数字我都提前给出
*避免了没必要的十分耗时的生成过程，大家也可以直接修改这些数字,即使是很大的数也支持！
*/
/*测试数据：
* 素数 p = 19, 本原元 g = 2, 私钥 x = 9, 公钥 y = 18, 随机取值 k = 5
*/
string prime("19"), prielem("2"), key("9"), pubKey("18"), randomK("5");
/*数的初始化*/
/*p 素数 g 本原元 x 私钥 y 公钥 k 随机取值*/
BigInteger p(prime, 10),g(prielem, 10), x(key, 10), y(pubKey, 10), k(randomK, 10),
one((__int64)1), two((__int64)2);
/*签名*/
void elgamalSign(unsigned char *message, int len, BigInteger &r, BigInteger &s)
{
/*m 为消息所对应的明文数值*/
unsigned char result[16] ={0};
md5(message, len, result);
/*输出MD5值*/
cout << "消息的MD5散列值为：" ;
for (int i = 0; i < 16; i++)
printf ("%02x", result[i]);
cout << endl;
/*用md5作为消息的hash值*/
/*用hash值初始化m*/
BigInteger m(result, 16), pMinusOne(p - one);
BigInteger k1;
r = g.modPow(k, p);
/*k1 为 k 在 p - 1 下的逆元*/
k1 = MultipInverse(k, pMinusOne);
s = ((m - x * r) * k1 ) % pMinusOne;
}
/*签名验证*/
bool elgamalVerifiSign(unsigned char *message, int len, BigInteger &r, BigInteger &s)
{
cout << "接收到消息的MD5散列值为：";
unsigned char result[16] ={0};
md5(message, len, result);
for (int i = 0; i < 16; i++)
printf ("%02x", result[i]);
cout << endl;
BigInteger leftValue, rightValue;
BigInteger m(result, 16);
leftValue = (y.modPow(r, p) * r.modPow(s, p)) % p;
rightValue = g.modPow(m, p);
if (leftValue == rightValue)
{
return true;
}
else
{
return false;
}
}
最后上一个主函数测试一下！
main.cpp
//本原元的概念：若模n下a的阶d=φ(n)，a就是n的本原元（又称为原根）。此时a是Z*_n的生成元。
/*======================================================
Diffie-Hellman 算法下面就给出一个快速求大素数 p 及其本原根的算法
算法如下：
P1. 利用素性验证算法，生成一个大素数 q；
P2. 令 p = q * 2 + 1；
P3. 利用素性验证算法，验证 p 是否是素数，如果 p 是合数，则跳转到 P1；
P4. 生成一个随机数 g，1 < g < p - 1；
P5. 验证 g2 mod p 和 gq mod p 都不等于 1，否则跳转到 P4；
P6. g 是大素数 p 的本原根。
======================================================*/
#include "ELGamal.h"
int main()
{
/*================================================================
//求一个大素数以及其本原元有点难度,速度慢到不行,我丢弃了这个想法,素数和本原元直接输入
BigInteger p((__int64)3), q, two((__int64)2), one((__int64)1), g, zero;
srand((unsigned)time(0));
bool flag = false;
while(!flag)
{
q = GetPrime(); //得到一个大素数
//cout << q << endl;
//system("pause");
p = q * two + one;
if(IsPrime(p))
flag = true;
}
while ((g * g) % p != one && (g * q) % p != one )
{
unsigned int len = rand() % 20;
for (int i = 0; i < len; i++) //产生一个随机数g
{
g.data[i] = (unsigned int)rand ();
if(g.data[i] != 0)
g.dataLength = i + 1;
}
}
cout << p << endl;
cout << g << endl;
========================================================*/
cout << "签名者 A:" << endl;
string message;
BigInteger r, s;
cout << "请输入要签名的消息：" << endl;
cin >> message;
elgamalSign((unsigned char *)message.c_str(), message.length(), r, s);
cout << "签名信息如下：" << endl;
/*不要奇怪为什么r总是等于d,去看一下r的定义就知道了。
*r = g^k mod p[g是mod p 下的本原元, k是任意取的常数(k 与 p - 1互素),p是素数]
*由于以上的这些数都提前给定了,所以结果也肯定是一个常数
*/
cout << "r = " << r << endl;
cout << "s = " << s << endl;
cout << endl;
/*
unsigned int len = rand() % 10;
for (int i = 0; i < len; i++)
{
k.data[i] = (unsigned int)rand ();
if(k.data[i] != 0)
k.dataLength = i + 1;
}
temp = p - one;
while(Gcd(k,p - one) != one)
{
k = k + two;
}
cout << k <<endl;
*/
cout << "现在将 消息 以及 r s 传递给接收方 B ~~~ ~~~" << endl;
cout << endl;
cout << "接受者 B:" << endl;
if (elgamalVerifiSign((unsigned char *)message.c_str(), message.length(), r, s))
{
cout << "签名有效" << endl;
}
else
{
cout << "签名无效" << endl;
}
system ("pause");
return 0;
}
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