大整数乘法（运算符重载） (25分)/高精算法

最新推荐文章于 2024-06-05 17:04:38 发布

原创最新推荐文章于 2024-06-05 17:04:38 发布 · 1.3k 阅读

7 ·

CC 4.0 BY-SA版权

opp 专栏收录该内容

7 篇文章

订阅专栏

BigInteger类表示不超过1000位的无符号大整数。试重载`>>`，`<<`和`*`，以支持无符号大整数的输入、输出与乘法。

重载面向BigInteger类对象的运算符：

>>
<<
*

裁判测试程序样例：

#include <iostream>
#include <string>
using namespace std;

/* 请在这里填写答案 */

int main() 
{
    BigInteger a, b, c;
    cin >> a >> b;
    c = a * b;
    cout << a << "*" << b << "=" << c << endl;
    return 0;
}

输入样例：

123456789
987654321

输出样例：

123456789*987654321=121932631112635269

代码实现：

没有满分的。。。

class BigInteger{
   public:
       BigInteger(){}
       friend ostream& operator<<(ostream &ou,BigInteger &a);
       friend istream& operator>>(istream &in,BigInteger &a);
       friend BigInteger operator*(BigInteger &a,BigInteger &b);
       string aa;
       int na[1002],re[2005],l;
};
ostream& operator<<(ostream &ou,BigInteger &a){
   for(int i=a.l-1;i>=0;i--)
       ou<<a.re[i];
   return ou;
}
istream& operator>>(istream &in,BigInteger &a){
   cin>>a.aa;
//   cout<<a.aa<<endl;
   a.l=a.aa.length();
   for(int i=0;i<a.l;i++)
       a.re[i]=a.na[i]=(a.aa[a.l-1-i]-'0');
   return in;
}
BigInteger operator*(BigInteger &a,BigInteger &b){
   BigInteger c;
   c.l=a.l+b.l;
   for(int i=0;i<a.l;i++)
       for(int j=0;j<b.l;j++)
           c.re[i+j]+=a.na[i]*b.na[j];
   while(c.re[c.l]==0)
       c.l--;
   for(int i=0;i<c.l;i++){
       c.re[i+1]+=c.re[i]/10;
       c.re[i]=c.re[i]%10;
   }
    if(c.re[c.l]!=0) c.l++;
   return c;
}

满分（借鉴别人的来纠错）

class BigInteger
{
private:
    string str;
public:
    BigInteger() {}
    BigInteger(string s)
    {
        str=s;
    }
    friend ostream&operator<<(ostream &os, BigInteger &bigint)
    {
        os<<bigint.str;
        return os;
    }
    friend istream&operator>>(istream &is,BigInteger &bigint)
    {
        is>>bigint.str;
        return is;
    }
    friend BigInteger operator*( BigInteger &bigint1, BigInteger &bigint2);
};
string reserve(string s)
{
    int l=s.length();
    int j=l-1;
    char temp;
    for(int i=0; i<l/2; i++,j--)
    {
        temp=s[i];
        s[i]=s[j];
        s[j]=temp;
    }
    return s;
}

BigInteger operator*( BigInteger &bigint1, BigInteger &bigint2)
{
    string a=reserve(bigint1.str);
    string b=reserve(bigint2.str);
    if(bigint1.str[0]=='0' || bigint2.str[0]=='0')
    {
        string st="0";
        return BigInteger(st);
    }
    int m,n;
    int *c=new int[a.length()+b.length()+2]();
    for(m=0; m<a.length(); m++)
    {
        for(n=0; n<b.length(); n++)
        {
            c[m+n]+=(a[m]-'0')*(b[n]-'0');
        }
    }

    m--;
    n--;
    int carry=0;
    char *ch=new char[a.length()+b.length()+5];
    int p;
    for( p=0; p<=m+n; p++)
    {
        if(c[p]>9)
        {
            ch[p]=(c[p]+carry)%10+'0';
            carry=(c[p]+carry)/10;
        }
        else
        {
            if(carry+c[p]>9)
            {
                ch[p]=(carry+c[p])%10+'0';
                carry=(carry+c[p])/10;
            }
            else
            {
                ch[p]=carry+c[p]+'0';
                carry=0;
            }
        }

    }
    if(carry>0)
    {
        ch[p]=carry+'0';
        ch[++p]='\0';
    }
    else
        ch[p]='\0';
    string st(ch);
    string s =reserve(st);
    return BigInteger(s);

}