#ifndef DSS_H

#define DSS_H

#include <string>

#include <iostream>

#include <cmath>

using namespace std;

class DSS

{

public:

unsigned __int64 x; //私钥

unsigned __int64 y; //公钥

unsigned __int64 r; //数字签名(r,s)

unsigned __int64 s;

string sig;

DSS() {

}

DSS(unsigned __int64 xy, string sigg) {

x = y = xy;

sig = sigg;

}

void signature(string message) {

//签名

srand(time(0));

hash<std::string> h;

unsigned __int64 k = rand() % q; //k为随机数 0<k<q

r = PowMod(g, k, p) % q;

s = mod_reverse(k, q)*((h(message) + x*r) % q) % q;

//r*1000+s 存入字符串sig

int str = r * 1000 + s;

char cnum[7];

sprintf_s(cnum, "%d", str);

sig = cnum;

while (sig.size() < 6) //若不够6位则在前面补0

{

sig = "0" + sig;

}

}

void verify(string message) {

//从sig里获取(x,s)

char cnum[7];

int str;

strcpy(cnum, sig.substr(0, 6).c_str());

sscanf(cnum, "%d", &str);

r = str / 1000;

s = str % 1000;

//验证

hash<std::string> h;

unsigned __int64 w = mod_reverse(s,q);

unsigned __int64 u1 = h(message) * w % q;

unsigned __int64 u2 = r * w % q;

unsigned __int64 v = (PowMod(g, u1, p)*PowMod(y, u2, p) % p) % q; //用了a*b % n = (a % n)*(b % n) % n

if (v == r) {

cout << "ture" << endl;

}

else

{

cout << "false" << endl;

}

}

//产生私钥x和公钥y

void generateKey() {

//0 < x < q;

srand(time(0));

x = rand() % q;

y = PowMod(g, x, p); //x为75时y为4567

cout << "私钥:" << x << endl;

cout << "公钥:" << y << endl;

}

private:

unsigned __int64 q = 101;

unsigned __int64 p = q * 78 + 1; // p = 7879

unsigned __int64 pMinus1Divideq = 78; // (p - 1) / q

unsigned __int64 h = 3;

unsigned __int64 g = PowMod(h, pMinus1Divideq, p); // g = 170

//模乘运算，返回值 x=a*b mod n

inline unsigned __int64 MulMod(unsigned __int64 a, unsigned __int64 b, unsigned __int64 n)

{

return a * b % n;

}

//模幂运算，返回值 x=base^pow mod n

unsigned __int64 PowMod(unsigned __int64 &base, unsigned __int64 &pow, unsigned __int64 &n)

{

unsigned __int64 a = base, b = pow, c = 1;

while (b)

{

while (!(b & 1))

{

b >>= 1; //a=a * a % n; //函数看起来可以处理64位的整数，但由于这里a*a在a>=2^32时已经造成了溢出，因此实际处理范围没有64位

a = MulMod(a, a, n);

} b--; //c=a * c % n; //这里也会溢出，若把64位整数拆为两个32位整数不知是否可以解决这个问题。

c = MulMod(a, c, n);

} return c;

}

//返回d=gcd(a,b);和对应于等式ax+by=d中的x,y

long long extend_gcd(long long a, long long b, long long &x, long long &y) {

if (a == 0 && b == 0) return -1;//无最大公约数

if (b == 0) {

x = 1; y = 0; return a;

}

long long d = extend_gcd(b, a%b, y, x);

y -= a / b*x;

return d;

}

//ax = 1(mod n) 求X 即: 求 1 / a % n

long long mod_reverse(long long a, long long n) {

long long x, y;

long long d = extend_gcd(a, n, x, y);

if (d == 1)

return (x%n + n) % n;

else return -1;

}

};

#endif