题解 | 高精度整数加法

#include <iostream>
using namespace std;
class strnum{
private:
  int hcf_dvqi(string& s1, string& s2){//-------------使两个数组一样长，并在前面加一个0防止越界，最后返回最终长度
    size_t size1 = s1.size(), size2 = s2.size();
    if (size1 < size2){
      for (int i = 0; i < size2 - size1; i++)
        s1 = '0' + s1;
    }
    if (size1 > size2)
      for (int i = 0; i < size1 - size2; i++)
        s2 = '0' + s2;
    s1 = '0' + s1;
    s2 = '0' + s2;
    return max(size1, size2) + 1;
  }
  void hcf_quiuqmdclk(string& ss){//-------------去除前导零
    while (ss[0] == '0' && ss.size() != 1)
      ss = ss.substr(1, ss.size() - 1);
  }
public:
  string s;
  strnum(string ss){
    s = ss;
  }
  string operator + (strnum& ss){//----------重载运算符‘+’
    int size = hcf_dvqi(s, ss.s);
    int rest = 0, temp;
    string ans = "";
    for (int i = size - 1; i >= 0; i--){
      temp = ((int)(s[i] + ss.s[i] - '0' - '0') + rest);
      rest = temp / 10;//--------进位用
      ans = (char)(temp % 10 + '0') + ans;
    }
    hcf_quiuqmdclk(ans);
    return ans;
  }
};
int main() {
  string s, ss;
  cin >> s >> ss;
  strnum s1(s), s2(ss);
  cout << s1 + s2;
}
// 64 位输出请用 printf("%lld")