高精度之重载运算符

• 这篇文章就纯属自娱自乐啦，受时间以及技术原因的限制，这里面的重载，减法支持的功能仅限于减一次（因为没有同是负号的判断），除法仅限于高精除单精（高精除高精的一个个的减或者二分实在是不想写了）……
• 结构体片段以及输出操作如下：

struct bignum
{
     int len,s[maxn];
     char flag;
     bignum()
     {
            len=1;
            flag=‘+‘;
            memset(s,0,sizeof(s));
     }
     bignum (int num)
     {
            *this=num;
     }
     bignum (const char *num)
     {
            *this=num;
     }
     bignum operator = (const char *a)
     {
            len=strlen(a);
            for (int i=1;i<=len;++i)
              s[i]=a[len-i]-‘0‘;
            return *this;
     }
     bignum operator = (const int num)
     {
            char a[maxn];
            sprintf(a,"%d",num);
            *this=a;
            return *this;
     }
     bignum operator + (const bignum &a)
     {
            bignum c;
            c.len=max(len,a.len)+1;
            for (int i=1;i<c.len;++i)
            {
                c.s[i]+=(s[i]+a.s[i]);
                c.s[i+1]+=c.s[i]/10;
                c.s[i]%=10;
            }
            if (c.s[c.len]==0)
              c.len--;
            return c;
     }
     bignum operator += (const bignum &a)
     {
            *this=*this+a;
            return *this;
     }
     bignum operator * (const bignum &a)
     {
            bignum c;
            c.len+=(len+a.len);
            for (int i=1;i<=len;++i)
              for (int j=1;j<=a.len;++j)
              {
                  c.s[i+j-1]+=(s[i]*a.s[j]);
                  c.s[i+j]+=(c.s[i+j-1]/10);
                  c.s[i+j-1]%=10;
              }
            while (c.s[c.len]==0)
               c.len--;
            return c;
     }
     bignum operator *= (const bignum &a)
     {
            *this=(*this) * a;
            return *this;
     }
     bool operator < (const bignum &a) const
     {
          if (len!=a.len)
            return len<a.len;
          for (int i=len;i>=1;--i)
            if (s[i]!=a.s[i])
              return s[i]<a.s[i];
          return false;
     }
     bool operator > (const bignum &a) const
     {
          return a<*this;
     }
     bool operator <= (const bignum &a) const
     {
          return !(*this>a);
     }
     bool operator >= (const bignum &a) const
     {
          return !(*this<a);
     }
     bool operator == (const bignum &a) const
     {
          return !((*this<a) || (*this>a));
     }
     bool operator != (const bignum &a) const
     {
          return !(*this==a);
     }
     void change (bignum &a,bignum &b)
     {
            bignum tmp=a;
            a=b;
            b=tmp;
     }
     bignum operator - (const bignum &a) const
     {
             bignum b=*this,c;
             if (b<a)
             {
                 c.flag=‘-‘;
                 c.len=a.len;
                 for (int i=1;i<=c.len;++i)
                 {
                     c.s[i]+=(a.s[i]-b.s[i]);
                     if (c.s[i]<0)
                     {
                         c.s[i]+=10;
                         c.s[i+1]-=1;
                     }
                 }
                 while (c.len==0)
                    c.len--;
                 return c;
             }
             c.len=b.len;
             for (int i=1;i<=c.len;++i)
             {
                 c.s[i]+=(b.s[i]-a.s[i]);
                 if (c.s[i]<0)
                 {
                      c.s[i]+=10;
                      c.s[i+1]-=1;
                 }
             }
             while (c.len==0)
               c.len--;
             return c;
     }
     bignum operator -= (const bignum &a)
     {
            *this=(*this)-a;
            return *this;
     }
     bignum operator / (const int n)
     {
            bignum c,b=*this;
            c.len=b.len;
            int x=0;
            for (int i=1;i<=n;++i)
            {
                c.s[i]=(x*10+b.s[i])/n;
                x=(x*10+b.s[i])%n;
            }
            while (c.s[c.len]==0)
               c.len--;
            return c;
     }
     bignum operator /= (const int a)
     {
            *this=*this/a;
            return *this;
     }
};

ostream& operator << (ostream &out,const bignum &x)
{
         for (int i=x.len;i>=1;--i)
           printf("%d",x.s[i]);
         return out;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。