HJ57 High precision integer addition ●●

HJ57 High precision integer addition ●●

describe

Enter two integers represented by the string str, and find the sum of the numbers represented by them.

Data range: $1\le len\left(str\right)\le 10000$

Enter a description:

Enter two strings. Ensure that the string contains only '0'~'9' characters

Output description:

Output the summed result

example

Input:
9876543210
1234567890
Output:
11111111100

answer

1. Simulation

Process the two long strings
from end to end, add them bit by bit, push the single digit result into the stack , and record the value of extra_add, add bit by bit again in the next cycle, and finally
pop the stack one by one.

Time complexity, space complexity: $O\left(n\right)$

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

int main()
{
    
    
    string s1, s2;
    while(cin >> s1 >> s2){
    
    
        int len1 = s1.length(), len2 = s2.length();      // 字符串长度
        int idx1 = len1-1, idx2 = len2-1;                // 指针下标
        int extra_add = 0, temp = 0;
        stack<char> st;        
        while(idx1 >= 0 || idx2 >= 0){
    
    
            int num1 = 0, num2 = 0;
            if(idx1 >= 0) num1 = s1[idx1--] - '0';       // 加数
            if(idx2 >= 0) num2 = s2[idx2--] - '0';       // 加数
            temp = extra_add + num1 + num2;              // 和 0~18
            if(temp >= 10){
    
    
                extra_add = 1;                           // 进位
            }else{
    
    
                extra_add = 0;
            }
            st.push(temp%10 + '0');                      // 个位数入栈
        }
        if(extra_add == 1) cout << '1';                  // 可能的进位
        while(!st.empty()){
    
                                  // 出栈输出
            cout << st.top();
            st.pop();
        }
    }
    return 0;
}