//二分探商优化 效率比短除法大大提高...
LongInt LongInt::operator / (const LongInt &obj) const{
assert(obj!=0 && "devide zero");
bool negative = (*this<0 and obj>0) or (*this>0 and obj<0);
LongInt a = this->getABS() , b = obj.getABS();
//两类特殊情况
if(a < b)
return 0;
if(a == b)
return negative ? -1 : 1;
LongInt lft = 1, rgt = a;
LongInt curSum = lft + a;
LongInt middle = curSum.divBy2();
LongInt one = 1;
//开始二分
while( not(a >= middle * b and a < (b + middle * b) )){
if(middle * b < a) //探商
lft = middle + one;
else
rgt = middle;
curSum = lft + rgt;
middle = curSum.divBy2();
}
return negative ? middle.getOpposite() : middle;
}
//各种关系运算符
bool LongInt::operator < (const LongInt &obj) const{
if(n.back()<0 and obj.n.back()>=0)
return true;
else if(n.back() >=0 and obj.n.back()<0)
return false;
else if(n.back() <0 and obj.n.back()<0 )
return !(this->getABS() < obj.getABS());//递归
else if(n.back() >= 0 and obj.n.back() >=0){
if(n.size() != obj.n.size())
return n.size() < obj.n.size();
for (int k = n.size() -1 ; k>=0; --k) {
if( n[k] != obj.n[k] )
return n[k] < obj.n[k];
}
}
return false;
}
LongInt LongInt::operator * (const LongInt& obj) const{
//乘法的处理 也要考虑不同的符号
LongInt result;
result.n.clear();
result = 0;
LongInt a = this->getABS();
LongInt b = obj.getABS();
bool negative = (*this<0 and obj>0) or (*this>0 and obj<0);
if (*this==0 or obj==0)
return 0;
for (int i=0;i<obj.n.size();++i)
{
LongInt midRes;//临时中间变量
midRes.n.clear();
if (i)//如果i大于0
for (int j=1;j<=i;++j)
midRes.n.push_back(0);
unsigned long long r = 0;//中间余数
for (int j = 0; ; ++j)
{
if (r == 0 && j >= n.size() )
break;
else{
unsigned long long t = r;
if (j < n.size()) t += (long long int)a.n[j] * b.n[i];
midRes.n.push_back(t % _base);
r = t / _base;
}
}
result += midRes;
}
return negative ? result.getOpposite() : result;
}
