Algorithm questions daily practice---Day 81: High-precision addition

Get into the habit of writing together! This is the 13th day of my participation in the "Nuggets Daily New Plan · April Update Challenge", click to view the event details .

1. Problem description

Given two positive integers (without leading zeros), compute their sum.

in, $1≤integer length≤100000$

Topic link: High-precision addition .

Second, the subject requirements

Example 1

输入： 12    23
输出： 35
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Example 2

输入： 123  98
输出： 221
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visit

1.高精度
2.建议用时15~30min
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3. Problem Analysis

The number of this question is obviously beyond the range of all shaping, we can use the string string storage, where the storage range of the string is: 1~65400个字符. After inputting through the string, we need to calculate the number, so the char->int type -'0'meets the .

We use an array to store each digit. Before the official start, we think about a question, is it stored in the forward direction or in the reverse direction?

For example, for the number 75842, which of the following storage methods should be used:

下标：0 1 2 3 4
数字：7 5 8 4 2

下标：4 3 2 1 0
数字：7 5 8 4 2
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Definitely use the second one! The first one, you can see if the highest position can move forward, and then it will be -1.

For the addition rule, use the corresponding bits learned in elementary school to add up, and add 1 to 10 when full.

Fourth, the encoding implementation

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
vector<int> add(vector<int> &a, vector<int> &b)  
{
    int i,k=0;//初始进位为0
    vector<int>c;
    for(i=0;i<a.size()||i<b.size();i++)//在范围内
    {
        if(i<a.size()) k+=a[i];
        if(i<b.size()) k+=b[i];
        c.push_back(k%10);//%10添加
        k=k/10;//进位
    }
    if(k)//确保遗留的进位添加进去
       c.push_back(k); 
    return c;
}
int main()
{
    string s1,s2;//字符串存储
    int i;
    cin>>s1>>s2;
    vector<int>a,b,c;
    for(i=s1.size()-1;i>=0;i--)//字符串转数字，倒序存储
        a.push_back(s1[i]-'0');
    for(i=s2.size()-1;i>=0;i--)//字符串转数字，倒序存储
        b.push_back(s2[i]-'0');
    c=add(a, b);//加法法则
    for (i=c.size()-1;i>=0;i--)//输出结果
        cout<<c[i];
    return 0;
}
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