大整数减法

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;

}