10106 - Product

高精度乘法题目，用了LRJ白书的模板。WA了许多次。

注意点：1.前导零需要过滤，应该写个函数（我直接偷懒用加法搞定了）

           2.乘法、加法的进位。

最后，模板不一定是最好的，适合自己的，自己写的出来的才是最好的。

最后再送几组测试用例

input: 
0 
2222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222
2 
9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
9239538994338789151251557336598603050049616822294053424980870425131464466182353830814916071789845042370142421852990439213459450627013510493542232079586565952646636598169299756958773885685805728065441653009173744078838973219505500897158270221204367245
9896336400014021081112872660026786571823539120528457013481728314902092185310581449120208631014862091874106743950514845492688570616578189552019550880999626199354705816824577956144319588556564026758407650715151019940083959555421056787710369504463191742
9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999

Ouput: 
0 
19999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999998
91437586069023901335141066474711666537718790039237144985051289985350175432524770084824378427555002818097099309346160871667571286412924312245148206827831443006000924957609921550337260262953945785774998477067644833241060308469618958625980383949243282674332743009843156607612702745027529447743579967907340959122060977535403916274356199174936757805300947243257666366727345740624830784342794119256500258114097942703603683608916604350586797991674579535204948315352029368384842883219960490930872682219290790
99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999980000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001

input：

000000000 0000000000000000
0000005    000000000005

output：

0

25

Input: 

4738787878 
87897987 
333 
333 
87897987 
4738787878 
00000000 
0 
1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111
2222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222

Output: 
416529915296201586 
110889 
416529915296201586 
0 
2469135802469135802469135802469135802469135802469135802469135802469135802469135802469135802469135802469

 

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
const int maxn=10000;
struct bign
{
  int len,s[maxn],flag;//flag 为表示正负备用 
  bign() {memset(s,0,sizeof(s));len=1;}
  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[maxn];
      sprintf(s,"%d",num);
      *this=s;
      return *this;
  }
  bign(int num){ *this=num; }
  bign(const char* num){ *this=num;}
  string str() const
  {
    string res="";
    for(int i=0;i<len;i++) res=(char)(s[i]+'0')+res;
    if(res=="") res="0";
    return res;
  }
  bign operator + (const bign& b) const
  {
      bign c;
      c.len=0;
      int max=0;
      if(len>b.len) max=len;
      else max=b.len;
      for(int i=0,g=0;g || i<max;i++)
      {
        int x=g;
        if(i<len) x+=s[i];
        if(i<b.len) x+= b.s[i];
        c.s[c.len++]=x%10;
        g=x/10;        
      }
      int flag=1;
      for(int i=c.len;i>=0;i--)
      {
        //cout<<"DEBUG:"<<i<<" "<<c.s[i]<<endl; 
        if(c.s[i]!=0)
        {
            flag=0;
            c.len=i+1;
        break;
        }
      }
      if(flag) c.len=0;
      //cout<<"DEBUG "<<str()<<"  "<<b.str()<<"  "<<c.str()<<endl;
      return c;
  }
  bign operator * (const bign& b) const
  {
      bign c,tmp=0;
      int x;
      for(int i=0;i<len;i++)
        for(int j=0;j<b.len;j++)
        {
            c.s[i+j]+=b.s[j]*s[i];
            if(c.s[i+j]>=10) 
            {
              c.s[i+j+1]+=c.s[i+j]/10;
              c.s[i+j]=c.s[i+j]%10;
            }
        }
      c.len=b.len+len+1;
      if(c.s[c.len]==0) c.len--;
      c=c+tmp;
      return c;
  }

};
istream& operator >> (istream &in, bign& x)
{
  string s;
  in >> s;
  x = s.c_str();
  return in;
}
ostream& operator << (ostream &out, const bign& x)
{
  out << x.str();
  return out;
}

int main()
{
  bign a,b;
  bign tmp=0;
  while(cin>>a>>b)
  {
      a=a+tmp;
      b=b+tmp;
      bign c=a*b;
      cout<<c<<endl;
  }
  
}