题解 | #高精度整数加法#

/*
思路：用字符串来模拟加法运算， 注意进位情况的考虑， 
    小技巧：短字符串前面补0的操作
*/

#include <algorithm>
#include <iostream>
#include <string>
using namespace std;

int main() {
    string str1, str2;
    cin >> str1 >> str2;
    int len1 = str1.length();
    int len2 = str2.length();
    // 两个整数位数不一致，短的在前面补0
    int diff = 0;
    if(len1 < len2){
        diff = len2 - len1;
        for(int i = 0; i < diff; i++){
            str1 = '0' + str1;
        }
    }else {
        diff = len1 - len2;
        for(int i = 0; i < diff; i++){
            str2 = '0' + str2;
        }
    }
    reverse(str1.begin(), str1.end());
    reverse(str2.begin(), str2.end());
    len1 = str1.length();
    len2 = str2.length();

    int carry = 0;
    string ans;
    int i = 0, j =0;
    for(; i < len1 && j < len2; i++,j++){
        int sum =  (str1[i] - '0') + (str2[j] - '0') + carry ;
        if( sum >= 10){
            carry = 1;
            ans += to_string(sum%10);
        }else{
            carry = 0;
            ans += to_string(sum);
        }
    }
    
    if(carry){
        ans += to_string(carry);
    }

    reverse(ans.begin(), ans.end());
    cout << ans << endl;
    return 0;
}
// 64 位输出请用 printf("%lld")