大整数减法

来源：互联网发布：手机数据恢复价格编辑：程序博客网时间：2024/04/29 16:38

自己花了几个小时写的

最后边界虽然是对的但是感觉还是很麻烦，等在想想或找到别人更好的再说吧

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

//字符串反转
void Reverse(char *str)
{
char* left = str;
char temp;

str = str+strlen(str)-1;
while(left < str)
{
  temp = *left;
  *left++ = *str;
  *str-- = temp;
}
}

void Bigsub(char* a,char* b,char* res)
{
int lena = strlen(a),
  lenb = strlen(b);
int max = lena >= lenb ? lena : lenb;
int min = lenb;
char *str = new char[max+1];
// 串组a内总是存着二者较长的串组
if(lena < lenb)
{
  min = lena;
  strcpy(str,a);
  strcpy(a,b);
  strcpy(b,str);
}

//cout<<a<<endl;
//cout<<b<<endl;
//开始计算
int x = 0,i,sub;
for(i = 0; i < max; i++)
{

  if(i < min)
   sub = (a[i] - '0') - (b[i] - '0') - x;
  else
   sub = (a[i] - '0') - x;

  //处理a[i] - b[i] < 0的情况，要在前一位接1,
  if(sub < 0)
  {
   res[i] = 10 + sub + '0';
   x = 1;
   //a[i+1] = a[i+1] - '0' - 1;

  }
  else
  {
   res[i] = sub + '0';
   x = 0;
  }
}

  if(res[i-1] == '0')
        res[i-1] = 0;
    else
       res[i] = 0;

}
int main()
{
char a[200],b[200],res[202];
cin>>a>>b;

Reverse(a);
Reverse(b);
Bigsub(a,b,res);

Reverse(res);
cout<<res<<endl;

return 0;
}