题解 | a+b 自定义大整形加法

//本题只要求两个高精度整数的加法，因此只需用到上面模板中关于构造、赋值、输入、输出和加法的部分即可
#include <iostream>
#include <cstdio>
#include <iterator>
#include <string>
#include <cstring>
using namespace std;

const int maxn= 10000;

struct biginteger{
    int digit[maxn];
    int length;
    biginteger(); //构造
    biginteger operator=(string str);   //赋值
    biginteger operator+(const biginteger& b);  //相加
    friend istream& operator>>(istream& in,biginteger& x);//输入  
    friend ostream& operator<<(ostream& out,biginteger& x);//输出
};

istream& operator>>(istream& in,biginteger& x){
    string str;
    in>>str;
    x=str;
    return in;
}

ostream& operator<<(ostream& out ,const biginteger& x){
    for(int i=x.length-1;i>=0;i--){
        out<<x.digit[i];
    }
    return out;
}

biginteger::biginteger(){
    memset(digit, 0, sizeof(digit)); //构造
    length = 0;
}
biginteger biginteger::operator=(string str){          //赋值
    memset(digit, 0, sizeof(digit));
    length = str.size();
    for (int i=0; i<length; i++) {
        digit[i]=str[length-i-1]-'0';
    }
    return *this;
}
biginteger biginteger::operator+(const biginteger& b){  //相加
    biginteger answer;
    int carry = 0;
    for(int i=0;i<length||i<b.length;i++){
        int current = carry + digit[i]+b.digit[i]; //和正常计算一样,先算当前位,再算进位
        carry = current/10;
        answer.digit[answer.length++] = current%10;  //将当前位存入答案
    }
    if(carry){
        answer.digit[answer.length++] = carry;
    }
    return answer;

}
int main(){
    biginteger a;
    biginteger b;
    while(cin>>a>>b){
        cout<<a+b<<endl;
    }
    return 0;
}