高精度加减乘运算
在看刘汝佳的入门白书，其中提到了高精度加法，上网参考了别人的做法后，写了高精度减法，高精度乘法，但除法仍没有解决。
本例高精度范围一千位的十进制大数加减乘。
基本结构：用大数组保留大数各个位上的数
关键点：
加法：保留本位和进位，处理方法本位保留之前进位的数后，若第一个数的对应位还有数(即没有超过最高位)，加对应位的数，第二个数同理；
bign是自定义结构体，最后面会给出完整结构
bign fun(bign temp)
{
bign sum;
sum.len = 0;
for ( int i = 0, high = 0; high || i < max(len, temp.len); i++)
{
int standard = high;
if (i < len)
standard += s[i];
if (i < temp.len)
standard += temp.s[i];
sum.s[sum.len++] = standard % 10;
high = standard / 10;
}
return sum;
}减法：主要是借位的问题比较麻烦，用循环找到借位点，再向后加10，但开始始终想保证位上为正。后来发现没有利用数据可以为负的特性，我们可以用对象位相减，若结果为负，对应位-1，后面一位加10，思路并不难。
bign ( bign temp)
{
bign result;
bign high;
bign low;
if(this->operator>(temp))
{
high = *this;
low = temp;
}
else
{
high = temp;
low = *this;
}
result.len = high.len;
for ( int i = 0; i < high.len; i++)
{
result.s[i] += high.s[i] - low.s[i];
if (result.s[i] < 0)
{
result.s[i] += 10;
result.s[i+1]--;
}
}
result.clean();
return result;
}乘法：模拟手工原理列竖式，对应位要移位（不是都从个位开始的），此时对应位存的不是一位数，再按照加法的思想，除10，和10取余。这里有一点需要注意，加减法的结果位数最大限度已知，而乘法位数不定，所以输出前要把前导0除去（预先用了很大的空间，全部置0了，所以结果前有很多0）
bign ( bign temp)
{
bign result;
result.len = len + temp.len;
for ( int i = 0; i < len; i++)
for ( int j = 0; j < temp.len; j++)
result.s[i+j] += s[i] * temp.s[j];
for ( int i = 0; i < result.len-1; i++)
{
result.s[i+1] += result.s[i] / 10;
result.s[i] = result.s[i] % 10;
}
result.clean();
return result;
}为了方便，把它写成一个模板，其中加入了重载运算符，作用可以使大数的结构像普通数据结构一样，进行
<< >> += -= *= < > <= >= !=的操作
下面完整代码及测试
#include
#include
#include
#define MAX 1000
using namespace std;
struct bign
{
int len, s[MAX];
bign()
{
memset(s, 0, sizeof(s));
len = 1;
};
void clean()
{
while (len > 1 && !s[len-1])
len--;
}
bign operator = (const char *num)
{
len = strlen(num);
for ( int i = 0; i < len; i++)
s[i] = num[len-i-1] - '0';
return *this;
}
bign operator = (int num)
{
char s[MAX];
sprintf(s, "%d", num);
*this = s;
return *this;
}
string str() const
{
string res = "";
for ( int i = 0; i < len; i++)
res = (char)(s[i] + '0') + res;
if (res == "")
res = "0";
return res;
}
bool end()
{
char s[100];
strcpy(s, str().c_str());
if (s[0] == '0')
return true;
else return false;
}
//加法
bign operator + (const bign &temp) const
{
bign sum;
sum.len = 0;
for ( int i = 0, high = 0; high || i < max(len, temp.len); i++)
{
int standard = high;
if (i < len)
standard += s[i];
if (i < temp.len)
standard += temp.s[i];
sum.s[sum.len++] = standard % 10;
high = standard / 10;
}
return sum;
}
//乘法
bign operator * (const bign &temp)
{
bign result;
result.len = len + temp.len;
for ( int i = 0; i < len; i++)
for ( int j = 0; j < temp.len; j++)
result.s[i+j] += s[i] * temp.s[j];
for ( int i = 0; i < result.len-1; i++)
{
result.s[i+1] += result.s[i] / 10;
result.s[i] = result.s[i] % 10;
}
result.clean();
return result;
}
//减法
bign operator - (const bign &temp)
{
bign result;
bign high;
bign low;
if(this->operator>(temp))
{
high = *this;
low = temp;
}
else
{
high = temp;
low = *this;
}
result.len = high.len;
for ( int i = 0; i < high.len; i++)
{
result.s[i] += high.s[i] - low.s[i];
if (result.s[i] < 0)
{
result.s[i] += 10;
result.s[i+1]--;
}
}
result.clean();
return result;
}
//除法
/* bign operator / (const bign &temp)
{
bign result;
if (this->operator<(temp))
{
result.operator=(0);
return result;
}
else (this)
}
*/
bign operator -= (const bign &temp)
{
*this = *this + temp;
return *this;
}
bign operator += (const bign &temp)
{
*this = *this + temp;
return *this;
}
bign operator *= (const bign &temp)
{
*this = *this * temp;
return *this;
}
bool operator < (const bign &temp) const
{
if (len != temp.len)
return len < temp.len;
for ( int i = len-1; i >= 0; i--)
if (s[i] != temp.s[i])
return s[i] < temp.s[i];
return false;
}
bool operator > (const bign &temp) const
{
return temp < *this;
}
bool operator <= (const bign &temp) const
{
return !(temp < *this);
}
bool operator >= (const bign &temp) const
{
return !(*this < temp);
}
bool operator != (const bign &temp) const
{
return temp < *this || temp > *this;
}
bool operator == (const bign &temp) const
{
return !(temp < *this) && !(*this < temp);
}
};
istream& operator >> (istream &in, bign &temp)
{
string s;
in >> s;
temp = s.c_str();
return in;
}
ostream& operator << (ostream &out, const bign &temp)
{
out << temp.str();
return out;
}
int main(int argc, char *argv[])
{
bign x,y;
while (cin >> x >> y)
{
cout << "x*y = " << y*x << endl;
cout << "x+y = " << x+y << endl;
cout << "x-y = "<< x-y << endl;
}
return 0;
}