【算法1-1】模拟与高精度—加法篇

同理参照高精度算法-减法篇，一个模板改过来的

解决跟高精度减法不同的问题：

 

没有前导零的存在，但是可能会出现位数增多的情况

比如：9999+9999=19998

 

if (ans[maxl + 1] > 0)
        maxl++;

 

 

 

终极代码：

#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
int arr1[200000], arr2[200000], ans[200000];
string jiafa(string s1, string s2)
{
    int maxl = max(s1.length(), s2.length());
    string ss;
    if (s1.length() < s2.length() || (s1.length() == s2.length() && s1 < s2))
    {
        return jiafa(s2, s1);
    }
    for (int i = 1; i <= s1.length(); i++)
        arr1[i] = s1[s1.size() - i] - '0';
    for (int i = 1; i <= s2.length(); i++)
        arr2[i] = s2[s2.size() - i] - '0';
    for (int i = 1; i <= maxl; i++)
        ans[i] = arr1[i] + arr2[i];
    for (int i = 1; i <=maxl; i++)
    {
        if (ans[i]>9)
        {
            ans[i] -= 10;
            ans[i + 1]++;
        }
    }
    if (ans[maxl + 1] > 0)
        maxl++;
        for (int i = maxl; i >0; i--)
            ss += ans[i] + '0';
    return ss;
}
int main()
{
    string s1, s2;
    cin >> s1 >> s2;
    cout << jiafa(s1, s2);
}