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#include<stdlib.h>
#include<time.h>
/*//大数运算

#define BI_MAXLEN 4
#define DEC 10
#define HEX 16

class CBigInt
{
public:
   int m_nSign;    //记录大数的符号,支持负值运算
   int m_nLength;  //记录0x10000000进制的位数,0-40之间,相当于2进制的0-1280位
   unsigned long m_ulvalue[BI_MAXLEN];   //记录每一位的“数字”

   CBigInt();
   ~CBigInt();

//将大数赋值为另一个大数    
   CBigInt& Mov(CBigInt& A);

//将大数赋值为编译器能够理解的任何整形常数或变量  
   CBigInt& Mov(unsigned __int64 A); 

//比较两个大数大小 
   int Cmp(CBigInt& A);

//计算两个大数的和 
   CBigInt Add(CBigInt& A);

//重载函数以支持大数与普通整数相加
   CBigInt Add(long A);

//计算两个大数的差
   CBigInt Sub(CBigInt& A);

//重载函数以支持大数与普通整数相减
   CBigInt Sub(long A);

//计算两个大数的积
   CBigInt Mul(CBigInt& A);

//重载函数以支持大数与普通整数相乘
   CBigInt Mul(long A);

//计算两个大数的商
   CBigInt Div(CBigInt& A);

//重载函数以支持大数与普通整数相除
   CBigInt Div(long A);

//计算两个大数相除的余数
   CBigInt Mod(CBigInt& A);

//重载函数以支持大数与普通整数相除求模
   long Mod(long A); 

//将输入的10进制或16进制字符串转换成大数
 //  int InPutFromStr(CString& str, const unsigned int system);

//将大数按10进制或16进制格式输出到字符串
 //  int OutPutToStr(CString& str, const unsigned int system);

//欧几里德算法求:Y=X.Euc(A),使满足:YX mod A = 1
   CBigInt Euc(CBigInt& A);

//蒙哥马利算法求:Y=X.Mon(A,B),使满足:X^A mod B = Y
   CBigInt Mon(CBigInt& A, CBigInt& B);
};


/*注意以上函数的声明格式,完全遵循普通整数运算的习惯,例如大数
Y=X+Z 相当于 Y.Mov(X.(Add(Z)),这样似乎没有Mov(Y,Add(X,Z))
看起来舒服,但是一旦我们重载运算符“=”为“Mov”,“+”为“Add”,
则Y.Mov(X.(Add(Z))的形式就等价于 Y=X+Z。

俺不知道其他编程语言里是否支持运算浮重载,至少这样定义函数格式
在C++里可以带来很大的方便。


//下面让我们来实现大数类的主要成员函数:

//初始化大数为0
CBigInt::CBigInt()
{
m_nSign=1;
m_nLength=1;
for(int i=0;i<BI_MAXLEN;i++)m_ulvalue[i]=0;
}

//采用缺省的解构函数
CBigInt::~CBigInt()
{
}

//大数比较,如果大数A位数比大数B多,当然A>B
//如果位数相同,则从高位开始比较,直到分出大小
int CBigInt::Cmp(CBigInt& A)
{
if(m_nLength>A.m_nLength)return 1;
if(m_nLength<A.m_nLength)return -1;
for(int i=m_nLength-1;i>=0;i--)
{
if(m_ulvalue[i]>A.m_ulvalue[i])return 1;
if(m_ulvalue[i]<A.m_ulvalue[i])return -1;
}
return 0;
}

//照搬参数的各属性
CBigInt& CBigInt::Mov(CBigInt& A)
{
m_nLength=A.m_nLength;
for(int i=0;i<BI_MAXLEN;i++)m_ulvalue[i]=A.m_ulvalue[i];
return *this;
}

//大数相加
//调用形式:N.Add(A),返回值:N+A
//若两大数符号相同,其值相加,否则改变参数符号再调用大数相减函数
/******************************************************************
/*例如:
     A  B  C
+       D  E
--------------
= S  F  G  H

其中,若C+E<=0xffffffff,则H=C+E,carry(进位标志)=0
     若C+E>0xffffffff,则H=C+E-0x100000000,carry=1

     若B+D+carry<=0xfffffff,则G=B+D,carry=0      
     若B+D+carry>0xfffffff,则G=B+D+carry-0x10000000,carry=1

     若carry=0,则F=A,S=0
     若carry=1,A<0xfffffff,则F=A+1,S=0
     若carry=1,A=0xfffffff,则F=0,S=1*/
/*****************************************************************
CBigInt CBigInt::Add(CBigInt& A)
{
CBigInt X;
if(X.m_nSign==A.m_nSign) 
{
X.Mov(*this);
int carry=0;
       unsigned __int64 sum=0;
       if(X.m_nLength<A.m_nLength)X.m_nLength=A.m_nLength;
for(int i=0;i<X.m_nLength;i++)
{
sum=A.m_ulvalue[i];
sum=sum+X.m_ulvalue[i]+carry;
X.m_ulvalue[i]=(unsigned long)sum;
if(sum>0xffffffff)carry=1;
else carry=0;
}
if(X.m_nLength<BI_MAXLEN)
{
X.m_ulvalue[X.m_nLength]=carry;
   X.m_nLength+=carry;
}
return X;
}
else{X.Mov(A);X.m_nSign=1-X.m_nSign;return Sub(X);}
}

//大数相减
//调用形式:N.Sub(A),返回值:N-A
//若两大数符号相同,其值相减,否则改变参数符号再调用大数相加函数
/******************************************************************
/*例如:
     A  B  C
-       D  E
--------------
=    F  G  H

其中,若C>=E,则H=C-E,carry(借位标志)=0
     若C<E,则H=C-E+0x100000000,carry=1

     若B-carry>=D,则G=B-carry-D,carry=0      
     若B-carry<D,则G=B-carry-D+0x10000000,carry=1

     若carry=0,则F=A
     若carry=1,A>1,则F=A-1
     若carry=1,A=1,则F=0*/
/*****************************************************************
CBigInt CBigInt::Sub(CBigInt& A)
{
CBigInt X;
if(m_nSign==A.m_nSign)
{
X.Mov(*this);
int cmp=X.Cmp(A); 
if(cmp==0){X.Mov(0);return X;}
int len,carry=0;
unsigned __int64 num;
unsigned long *s,*d;
       if(cmp>0)
               {
                       s=X.m_ulvalue;
                       d=A.m_ulvalue;
                       len=X.m_nLength;
               }
       if(cmp<0)
               { 
                       s=A.m_ulvalue;
                       d=X.m_ulvalue;
                       len=A.m_nLength;
                       X.m_nSign=1-X.m_nSign;
               }
       for(int i=0;i<len;i++)
{
if((s[i]-carry)>=d[i])
{
X.m_ulvalue[i]=s[i]-carry-d[i];
carry=0;
}
else
{
num=0x100000000+s[i];
X.m_ulvalue[i]=(unsigned long)(num-carry-d[i]);
carry=1;
}
}
while(X.m_ulvalue[len-1]==0)len--;
X.m_nLength=len;
return X;
}
else{X.Mov(A);X.m_nSign=1-X.m_nSign;return Add(X);}
}

//大数相乘
//调用形式:N.Mul(A),返回值:N*A
/******************************************************************/
/*例如:
        A  B  C
*          D  E
----------------
=    S  F  G  H
+ T  I  J  K
----------------
= U  V  L  M  N

其中,SFGH=ABC*E,TIJK=ABC*D

而对于:
     A  B  C
*          E
-------------
= S  F  G  H    

其中,若C*E<=0xffffffff,则H=C*E,carry(进位标志)=0
     若C*E>0xffffffff,则H=(C*E)&0xffffffff
       carry=(C*E)/0xffffffff
     若B*E+carry<=0xffffffff,则G=B*E+carry,carry=0
     若B*E+carry>0xffffffff,则G=(B*E+carry)&0xffffffff
       carry=(B*E+carry)/0xffffffff
     若A*E+carry<=0xffffffff,则F=A*E+carry,carry=0
     若A*E+carry>0xffffffff,则F=(A*E+carry)&0xffffffff
       carry=(A*E+carry)/0xffffffff
     S=carry*/
/*****************************************************************
CBigInt CBigInt::Mul(CBigInt& A)
{
CBigInt X,Y;
unsigned __int64 mul;
       unsigned long carry;
       for(int i=0;i<A.m_nLength;i++)
{
Y.m_nLength=m_nLength;
carry=0;
for(int j=0;j<m_nLength;j++)
{
mul=m_ulvalue[j];
mul=mul*A.m_ulvalue[i]+carry;
Y.m_ulvalue[j]=(unsigned long)mul;
carry=(unsigned long)(mul>>32);
}
if(carry&&(Y.m_nLength<BI_MAXLEN))
               {
                       Y.m_nLength++;
                       Y.m_ulvalue[Y.m_nLength-1]=carry;
               }
if(Y.m_nLength<BI_MAXLEN-i)
{
Y.m_nLength+=i;
       for(int k=Y.m_nLength-1;k>=i;k--)Y.m_ulvalue[k]=Y.m_ulvalue[k-i];
       for(k=0;k<i;k++)Y.m_ulvalue[k]=0;
}
X.Mov(X.Add(Y));
}
if(m_nSign+A.m_nSign==1)X.m_nSign=0;
else X.m_nSign=1;
return X;
}

//大数相除
//调用形式:N.Div(A),返回值:N/A
//除法的关键在于“试商”,然后就变成了乘法和减法
//这里将被除数与除数的试商转化成了被除数最高位与除数最高位的试商
CBigInt CBigInt::Div(CBigInt& A)
{
CBigInt X,Y,Z;
int len;
unsigned __int64 num,div;
unsigned long carry=0;
Y.Mov(*this);
while(Y.Cmp(A)>0)
{       
if(Y.m_ulvalue[Y.m_nLength-1]>A.m_ulvalue[A.m_nLength-1])
{
len=Y.m_nLength-A.m_nLength;
div=Y.m_ulvalue[Y.m_nLength-1]/(A.m_ulvalue[A.m_nLength-1]+1);//高位试除
}
else if(Y.m_nLength>A.m_nLength)
{
len=Y.m_nLength-A.m_nLength-1;
num=Y.m_ulvalue[Y.m_nLength-1];
num=(num<<32)+Y.m_ulvalue[Y.m_nLength-2];
if(A.m_ulvalue[A.m_nLength-1]==0xffffffff)div=(num>>32);
else div=num/(A.m_ulvalue[A.m_nLength-1]+1);
}
else
{
                       X.Mov(X.Add(1));//应去掉???????????????
break;
}
               Z.Mov(div);
Z.m_nLength+=len;
for(int i=Z.m_nLength-1;i>=len;i--)Z.m_ulvalue[i]=Z.m_ulvalue[i-len];
for(i=0;i<len;i++)Z.m_ulvalue[i]=0;
X.Mov(X.Add(Z));
Z.Mov(Z.Mul(A));
Y.Mov(Y.Sub(Z));
}
if(Y.Cmp(A)==0)X.Mov(X.Add(1));
if(m_nSign+A.m_nSign==1)X.m_nSign=0;
else X.m_nSign=1;
return X;
}

//大数求模
//调用形式:N.Mod(A),返回值:N%A
//求模与求商原理相同
CBigInt CBigInt::Mod(CBigInt& A)
{
CBigInt X,Y;
int len;
unsigned __int64 num,div;
unsigned long carry=0;
X.Mov(*this);
while(X.Cmp(A)>0)
{       
if(X.m_ulvalue[X.m_nLength-1]>A.m_ulvalue[A.m_nLength-1])
{
len=X.m_nLength-A.m_nLength;
div=X.m_ulvalue[X.m_nLength-1]/(A.m_ulvalue[A.m_nLength-1]+1);
}
else if(X.m_nLength>A.m_nLength)
{
len=X.m_nLength-A.m_nLength-1;
num=X.m_ulvalue[X.m_nLength-1];
num=(num<<32)+X.m_ulvalue[X.m_nLength-2];
if(A.m_ulvalue[A.m_nLength-1]==0xffffffff)div=(num>>32);
else div=num/(A.m_ulvalue[A.m_nLength-1]+1);
}
else
{
X.Mov(X.Sub(A));//应去掉???????????????
break;
}
               Y.Mov(div);
Y.Mov(Y.Mul(A));
Y.m_nLength+=len;
for(int i=Y.m_nLength-1;i>=len;i--)Y.m_ulvalue[i]=Y.m_ulvalue[i-len];
for(i=0;i<len;i++)Y.m_ulvalue[i]=0;
X.Mov(X.Sub(Y));
}
if(X.Cmp(A)==0)X.Mov(0);
return X;
}


//暂时只给出了十进制字符串的转化
/*int CBigInt::InPutFromStr(CString& str, const unsigned int system=DEC)
{
       int len=str.GetLength();
Mov(0);
for(int i=0;i<len;i++)
       {
             Mov(Mul(system));
int k=str[i]-48;
Mov(Add(k));
 }
 return 0;
}

//暂时只给出了十进制字符串的转化
int CBigInt::OutPutToStr(CString& str, const unsigned int system=DEC)
{
str="";
char ch;
CBigInt X;
X.Mov(*this);
while(X.m_ulvalue[X.m_nLength-1]>0)
{
ch=X.Mod(system)+48;
str.Insert(0,ch);
       X.Mov(X.Div(system));
}
return 0;
}

//欧几里德算法求:Y=X.Euc(A),使满足:YX mod A=1
//相当于对不定方程ax-by=1求最小整数解
//实际上就是初中学过的辗转相除法
/********************************************************************
/*例如:11x-49y=1,求x

           11 x  -  49 y  =   1      a)
49%11=5 ->  11 x  -   5 y  =   1      b)
11%5 =1 ->     x  -   5 y  =   1      c)

令y=1  代入c)式  得x=6
令x=6  代入b)式  得y=13
令y=13 代入a)式  得x=58
/********************************************************************
CBigInt CBigInt::Euc(CBigInt& A)
{
CBigInt X,Y;
X.Mov(*this);
Y.Mov(A);
if((X.m_nLength==1)&&(X.m_ulvalue[0]==1))return X;
if((Y.m_nLength==1)&&(Y.m_ulvalue[0]==1)){X.Mov(X.Sub(1));return X;}//?????????
if(X.Cmp(Y)==1)X.Mov(X.Mod(Y));
else Y.Mov(Y.Mod(X));
X.Mov(X.Euc(Y));
       Y.Mov(*this);
if(Y.Cmp(A)==1)
{
X.Mov(X.Mul(Y));
X.Mov(X.Sub(1));
X.Mov(X.Div(A));
}
else
{
X.Mov(X.Mul(A));
X.Mov(X.Add(1));
X.Mov(X.Div(Y));
}
return X;
}

//蒙哥马利算法求:Y=X.Mon(A,B),使满足:X^A mod B=Y
//俺估计就是高中学过的反复平方法
CBigInt CBigInt::Mon(CBigInt& A, CBigInt& B)
{
CBigInt X,Y,Z;
X.Mov(1);
Y.Mov(*this);
       Z.Mov(A);
while((Z.m_nLength!=1)||Z.m_ulvalue[0])
{
if(Z.m_ulvalue[0]&1)
{
Z.Mov(Z.Sub(1));
X.Mov(X.Mul(Y));
X.Mov(X.Mod(B));
}
else
{
Z.Mov(Z.Div(2));
Y.Mov(Y.Mul(Y));
Y.Mov(Y.Mod(B));
}
}
       return X;
}*/
//计算u模m的逆
int MODR(int u, int m)
{
   int r1 = m, r2 = u, b1 = 0, b2 = 1;
   int q = r1/r2, r3 = r1%r2;
   int b3 = b1 - q*b2;
   while(r3)
   {
      r1 = r2; r2 = r3;
      b1 = b2; b2 = b3;
      q = r1/r2; r3 = r1%r2;
      b3 = b1 - q*b2;
   }
   return b2;
}
//计算x^r mod n
__int64 MOD(__int64 x, __int64 r, __int64 n)
{
	__int64 a = x, b = r, c = 1;
	while(b!=0)
	{cout<<"::::::::::::::::::::";
	     unsigned long p;
   p = b/0x100000000;
   cout<<p;
   p = b%0x100000000;
   cout<<p<<endl;      
		if(b%2)
		{
			b = b - 1;
		    c = (c*a)%n;
		}
		else
		{
			b = b/2;
			a = (a*a)%n;
		}	
	}
	return c;
}
//产生64位随机数
/*__int64 Random64()
{
	__int64 random = 0;
	srand((unsigned)time(NULL));
	for(int i=0; i<4; i++)
	{
        random = random<<16;//左移16位,生成高字位   
        random = random | rand();//或16位随机数
	}
	return random;
}*/
__int64 Random()
{
	__int64 random = 0;
	srand((unsigned)time(NULL));
	for(int i = 0; i<64; i++)
	{
		random = random<<1;
		random = random | rand()%2;
	}
	return random;
}
//Robin-Miller素性检验,产生64位的素数
__int64 Generate_Prime()
{
	__int64 prime;
	__int64 n, m, a, r, b =0;  
	int K = 10;
	while(1)  
	{
		do
		{
			prime = Random();
		}while(prime%2 == 0);//产生64位奇随机数
		n = prime - 1;
		while(n%2 == 0)   //n=2^b * m;
		{
			n = n/2;
			b++;
		}
		m = n;
		for(int t = 0; t < K; t++)//进行K次素性检验
		{
			//选择小于prime的随机数
			do
			{
				a = Random();
			}while(a >= prime);
			cout<<"::::::::::::::::::::"<<endl;
			r = MOD(a, m ,prime);	
			if(r!=1 && r!=prime-1)
			{
				for(int i = 1; i < b; i++)  //未通过检验则继续
				{
					r = (r * r) % prime;
					if(r == prime-1) break;
				}
				if(i == b) break;//未通过检验
			}
		}
		if(t == K) //通过检验则返回素数
			return prime;
	}
}
 
//判断互素
bool Husu(int a, int b)
{
	int c;
	if(a<b)
	{
		c = a; a = b; b = c;
	}
	while(b)
	{
		c = b;
		b = a%b;
		a = c;
		if(b ==1)
			return true;
	}
	return false;
}

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