高精度四则运算之加法实现——参考ACM/ICPC程序设计与分析(c++实现)chapter4

#include <cstdlib>
#include <iostream>
#include <algorithm>

using namespace std;
const int MAXLEN=100;

int*Str2Int(char*str)
{
  int i,len=strlen(str);
  int*a=new int[(len+1)*sizeof(int)];
  for(i=0;i<len;i++) a[i]=(int)str[len-i-1]-48;
  return a;
}

int check(int* a,int n)//规整 
{
   int k=0,len=n;
   while(a[len-1]==0&&len>1) len--;
   for(k=0;k<len;k++)
   {
      if(a[k]>=10)
      {
          a[k+1]=a[k+1]+a[k]/10;
          a[k]=a[k]%10;
      } 
   
   }
   if(a[k]!=0) len=k+1;
   return len;
    
}

char* Int2Str(int* a,int n)
{
   int i;
   char* str=new char[(n+1)*sizeof(char)];
   for(i=0;i<n;i++) str[i]=(char)a[n-i-1]+48;
   str[n]='';
   return str;

} 



//加法：输入两个用字符数组表示的长整数m1,m2
//返回一个字符数组指针，指向m1+m2的和
char* addition(char* m1,char*m2)
{
int i,len1,len2,len,c=0;
int*t1,*t2;
len1=strlen(m1);len2=strlen(m2);
len=max(len1,len2);
t1=new int[(len+2)*sizeof(int)]; 
t2=new int[(len+2)*sizeof(int)];
t1=Str2Int(m1);
t2=Str2Int(m2);
for(i=len1;i<len+1;i++) t1[i]=0;
for(i=len2;i<len+1;i++) t2[i]=0;
for(i=0;i<len;i++) t1[i]=t1[i]+t2[i];
len=check(t1,len);
return Int2Str(t1,len);
} 



int main(int argc, char *argv[])
{ char s1[MAXLEN],s2[MAXLEN];
 while(cin>>s1>>s2)
 cout<<addition(s1,s2)<<endl;
    
    system("PAUSE");
    return EXIT_SUCCESS;
}