//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;
}
LongInt LongInt::operator - (const LongInt& obj) const{
LongInt result; result.n.clear();
if(*this == obj)
result = 0;
else{
//两个都是非负数时
if(*this >=0 and obj >= 0){
if(*this > obj){
for (int i = 0, r = 0;; ++i) {
if(not r and i>= n.size() and i>= obj.n.size())
break;
int t = r;
if(i < n.size())
t += n[i];
if(i < obj.n.size())
t -= obj.n[i];
if(t < 0){
t += _base; r = -1; //进位
}else{
r = 0;
}
result.n.push_back(t);
}
//除去前位0
while(not result.n.back() and not result.n.empty())
result.n.pop_back();
if(result.n.empty())
result.n.push_back(0);
//如果就是0 则补回一个
}else{
result = (obj - *this).getOpposite();
}
}
else if(*this>=0 and obj<0)
result = *this + obj;
else if(*this<0 and obj>=0)
result = (this->getOpposite() + obj).getOpposite();
else if(*this<0 and obj<0)
//递归即可
result = obj.getOpposite() - this->getOpposite();
}
return result;
}
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;
}
//各种算法运算符
LongInt LongInt::operator + (const LongInt& obj) const{
LongInt result;
result.n.clear();
if(*this >= 0 and obj >= 0){//两个非负整数相加才是真正的相加 否则转移为减法处理
for (int i = 0, r = 0; ; ++i) {
if(r==0 and i>=n.size() and i>=obj.n.size())
break;
int t = r;
if(i < n.size())
t += n[i];
if(i < obj.n.size())
t += obj.n[i];
result.n.push_back(t % _base);
r = t / _base;
}
}else if(*this < 0 and obj < 0)
result = (this->getABS() + this->getABS()).getOpposite();
else if(*this < 0 and obj >= 0)
result = obj - this->getOpposite();
else if(*this >= 0 and obj < 0)
result = *this - obj.getOpposite();
return result;
}
